Binary Tress-Simplified
FAQ on Tress
“ A binary tree is a tree data structure in which each node has at most two children, which are referred to as the left child and the right child.”
A recursive definition using just set theory notions is that a (non-empty) binary tree is a tuple (L, S, R), where L and R are binary trees or the empty set and S is a singleton set”.
Let’s quickly talk about tree terminology.
- A tree is a data structure that consists of one or more nodes organized into a hierarchy.
- The tree has one root, which is the top node.
- All nodes except the root have a unique parent.
- The node labeled * in the picture below is a parent. Nodes labeled 2 and 7 are its children; children are ordered from left to right.
- A node with no children is called a leaf node.
- A node that has one or more children and that is not the root is called an interior node.
- The children can also be complete subtrees. In the picture above the left child (labeled *) of the + node is a complete subtree with its own children.
- In computer science we draw trees upside down starting with the root node at the top and branches growing downward.
Binary Search Tree (BST)
“Binary search tree” (BST) or “ordered binary tree” is a type of binary tree where the nodes are arranged in order:
For each node, all elements in its left subtree are less-or-equal to the node (<=), and all the elements in its right subtree are greater than the node (>).
The tree shown above is a binary search tree —
the “root” node is a 5, and its left subtree nodes (1, 3, 4) are <= 5, and it’s right subtree nodes (6, 9) are > 5.
Recursively, each of the subtrees must also obey the binary search tree constraint: in the (1, 3, 4) subtree, the 3 is the root, the 1 <= 3 and 4 > 3.
The nodes at the bottom edge of the tree have empty subtrees and are called “leaf” nodes (1, 4, 6) while the others are “internal” nodes (3, 5, 9).
Binary Search Tree Application:
Basically, binary search trees are fast at insert and lookup. The next section presents the code for these two algorithms. On average, a binary search tree algorithm can locate a node in an N node tree in order lg(N) time (log base 2).
Therefore, binary search trees are good for “dictionary” problems where the code inserts and looks up information indexed by some key. The lg(N) behavior is the average case — it’s possible for a particular tree to be much slower depending on its shape.
Typical Binary Tree Code in C/C++
As an introduction, we’ll look at the code for the two most basic binary search tree operations — lookup() and insert(). The code here works for C or C++.
The binary tree is built with a node type like this…
struct node {
int data;
struct node* left;
struct node* right;
}
/*Helper function that allocates a new node with the given data and NULL left and right pointers.*/
struct node* NewNode(int data) {
struct node* node = new(struct node); // “new” is like “malloc”
node->data = data;
node->left = NULL;
node->right = NULL;
return(node);
}
/*Given a binary tree, return true if a node with the target data is found in the tree.*/
static int lookup(struct node* node, int target) {
// 1. Base case == empty tree
// in that case, the target is not found so return false
if (node == NULL) {
return(false);
}
else {
// 2. see if found here
if (target == node->data) return(true);
else {
// 3. otherwise recur down the correct subtree
if (target < node->data) return(lookup(node->left, target));
else return(lookup(node->right, target));
}
}
}
/* Given a binary search tree and a number, inserts a new node with the given number in the correct place in the tree. Returns the new root pointer which the caller should then use (the standard trick to avoid using reference parameters).*/
struct node* insert(struct node* node, int data) {
// 1. If the tree is empty, return a new, single node
if (node == NULL) {
return(newNode(data));
}
else {
// 2. Otherwise, recur down the tree
if (data <= node->data) node->left = insert(node->left, data);
else node->right = insert(node->right, data);
return(node); // return the (unchanged) node pointer
}
}
The shape of a binary tree depends very much on the order that the nodes are inserted. In particular, if the nodes are inserted in increasing order (1, 2, 3, 4), the tree nodes just grow to the right leading to a linked list shape where all the left pointers are NULL. A similar thing happens if the nodes are inserted in decreasing order (4, 3, 2, 1).
With the basic methods in place let’s solve the below frequently asked problems on trees.
Solving BST made simple:
Write code that builds the following little 1–2–3 binary search tree.
2
/ \
1 3
Write the code in three different ways…
- a: by calling newNode() three times, and using three pointer variables
- b: by calling newNode() three times, and using only one pointer variable
- c: by calling insert() three times passing it the root pointer to build up the tree
SOLUTION:
1)By calling newNode() three times, and using three pointer variables
struct node* build123a() {
struct node* root = newNode(2);
struct node* lChild = newNode(1);
struct node* rChild = newNode(3);
root->left = lChild;
root->right= rChild;
return(root);
}
2)call newNode() three times, and use only one local variable
struct node* build123b() {
struct node* root = newNode(2);
root->left = newNode(1);
root->right = newNode(3);
return(root);
}
3)Build 123 by calling insert() three times.
struct node* build123c() {
struct node* root = NULL;
root = insert(root, 2);
root = insert(root, 1);
root = insert(root, 3);
return(root);
}
PROBLEM
Demonstrate simple binary tree traversal. Given a binary tree, count the number of nodes in the tree.
int size(struct node* node) {
if (node==NULL) {
return(0);
} else {
return(size(node->left) + 1 + size(node->right));
}
}
PROBLEM
Compute the “maxDepth” of a tree — the number of nodes along the longest path from the root node down to the farthest leaf node.
int maxDepth(struct node* node) {
if (node==NULL) {
return(0);
}
else {
// compute the depth of each subtree
int lDepth = maxDepth(node->left);
int rDepth = maxDepth(node->right);
// use the larger one
if (lDepth > rDepth) return(lDepth+1);
else return(rDepth+1);
}
}
PROBLEM
Given a non-empty binary search tree, return the minimum data value found in that tree.Note that the entire tree does not need to be searched.
int minValue(struct node* node) {
struct node* current = node;
// loop down to find the leftmost leaf
while (current->left != NULL) {
current = current->left;
}
return(current->data);
}
PROBLEM
Given a binary search tree, print out its data elements in increasing sorted order.
void printTree(struct node* node) {
if (node == NULL) return;
printTree(node->left);
printf(“%d “, node->data);
printTree(node->right);
}
PROBLEM
Given a binary tree, print its nodes according to the “bottom-up” postorder traversal.
void printPostorder(struct node* node) {
if (node == NULL) return;
// first recur on both subtrees
printTree(node->left);
printTree(node->right);
// then deal with the node
printf(“%d “, node->data);
}
PROBLEM:
Given a tree and a sum, return true if there is a path from the root down to a leaf, such that adding up all the values along the path equals the given sum.
Strategy: subtract the node value from the sum when recurring down, and check to see if the sum is 0 when you run out of tree.
int hasPathSum(struct node* node, int sum) {
// return true if we run out of tree and sum==0
if (node == NULL) {
return(sum == 0);
}
else {
// otherwise check both subtrees
int subSum = sum — node->data;
return(hasPathSum(node->left, subSum) ||
hasPathSum(node->right, subSum));
}
}
PROBLEM
Given a binary tree, print out all of its root-to-leaf paths, one per line.
void printPaths(struct node* node) {
int path[1000];
printPathsRecur(node, path, 0);
}
/*
Recursive helper function — given a node, and an array containing
the path from the root node up to but not including this node,
print out all the root-leaf paths.
*/
void printPathsRecur(struct node* node, int path[], int pathLen) {
if (node==NULL) return;
// append this node to the path array
path[pathLen] = node->data;
pathLen++;
// it’s a leaf, so print the path that led to here
if (node->left==NULL && node->right==NULL) {
printArray(path, pathLen);
}
else {
// otherwise try both subtrees
printPathsRecur(node->left, path, pathLen);
printPathsRecur(node->right, path, pathLen);
}
}
// Utility that prints out an array on a line.
void printArray(int ints[], int len) {
int i;
for (i=0; i<len; i++) {
printf(“%d “, ints[i]);
}
printf(“\n”);
}
PROBLEM:
Change a tree so that the roles of the left and right pointers are swapped at every node.
So the tree…
4
/ \
2 5
/ \
1 3
is changed to…
4
/ \
5 2
/ \
3 1
void mirror(struct node* node) {
if (node==NULL) {
return;
}
else {
struct node* temp;
// do the subtrees
mirror(node->left);
mirror(node->right);
// swap the pointers in this node
temp = node->left;
node->left = node->right;
node->right = temp;
}
}
PROBLEM
For each node in a binary search tree, create a new duplicate node, and insert the duplicate as the left child of the original node.
The resulting tree should still be a binary search tree.
So the tree…
2
/ \
1 3
Is changed to…
2
/ \
2 3
/ /
1 3
/
1
void doubleTree(struct node* node) {
struct node* oldLeft;
if (node==NULL) return;
// do the subtrees
doubleTree(node->left);
doubleTree(node->right);
// duplicate this node to its left
oldLeft = node->left;
node->left = newNode(node->data);
node->left->left = oldLeft;
}
PROBLEM
Given two trees, return true if they are structurally identical.
int sameTree(struct node* a, struct node* b) {
// 1. both empty -> true
if (a==NULL && b==NULL) return(true);
// 2. both non-empty -> compare them
else if (a!=NULL && b!=NULL) {
return(
a->data == b->data &&
sameTree(a->left, b->left) &&
sameTree(a->right, b->right)
);
}
// 3. one empty, one not -> false
else return(false);
}
PROBLEM
For the key values 1…numKeys, how many structurally unique binary search trees are possible that store those keys.
Strategy: consider that each value could be the root. Recursively find the size of the left and right subtrees.
int countTrees(int numKeys) {
if (numKeys <=1) {
return(1);
}
else {
// there will be one value at the root, with whatever remains
// on the left and right each forming their own subtrees.
// Iterate through all the values that could be the root…
int sum = 0;
int left, right, root;
for (root=1; root<=numKeys; root++) {
left = countTrees(root — 1);
right = countTrees(numKeys — root);
// number of possible trees with this root == left*right
sum += left*right;
}
return(sum);
}
}
PROBLEM
Return true if a binary tree is a binary search tree.
int isBST(struct node* node) {
if (node==NULL) return(true);
// false if the max of the left is > than us
// (bug — an earlier version had min/max backwards here)
if (node->left!=NULL && maxValue(node->left) > node->data)
return(false);
// false if the min of the right is <= than us
if (node->right!=NULL && minValue(node->right) <= node->data)
return(false);
// false if, recursively, the left or right is not a BST
if (!isBST(node->left) || !isBST(node->right))
return(false);
// passing all that, it’s a BST
return(true);
}
PROBLEM
Returns true if the given tree is a binary search tree (efficient version).
int isBST2(struct node* node) {
return(isBSTUtil(node, INT_MIN, INT_MAX));
}
/*
Returns true if the given tree is a BST and its
values are >= min and <= max.
*/
int isBSTUtil(struct node* node, int min, int max) {
if (node==NULL) return(true);
// false if this node violates the min/max constraint
if (node->data<min || node->data>max) return(false);
// otherwise check the subtrees recursively,
// tightening the min or max constraint
return
isBSTUtil(node->left, min, node->data) && isBSTUtil(node->right, node->data+1, max)
);
}
Java Binary Trees and Solutions
In Java, the key points in the recursion are exactly the same as in C or C++. In fact, I created the Java solutions by just copying the C solutions, and then making the syntactic changes. The recursion is the same, however the outer structure is slightly different.
In Java, we will have a BinaryTree object that contains a single root pointer. The root pointer points to an internal Node class that behaves just like the node struct in the C/C++ version. The Node class is private — it is used only for internal storage inside the BinaryTree and is not exposed to clients. With this OOP structure, almost every operation has two methods: a one-line method on the BinaryTree that starts the computation, and a recursive method that works on the Node objects. For the lookup() operation, there is a BinaryTree.lookup() method that the client uses to start a lookup operation. Internal to the BinaryTree class, there is a private recursive lookup(Node) method that implements the recursion down the Node structure. This second, private recursive method is basically the same as the recursive C/C++ functions above — it takes a Node argument and uses recursion to iterate over the pointer structure.
Java Binary Tree Structure
To get started, here are the basic definitions for the Java BinaryTree class, and the lookup() and insert() methods as examples…
// BinaryTree.java
public class BinaryTree {
// Root node pointer. Will be null for an empty tree.
private Node root;
/*
— Node —
The binary tree is built using this nested node class.
Each node stores one data element, and has left and right
sub-tree pointer which may be null.
The node is a “dumb” nested class — we just use it for
storage; it does not have any methods.
*/
private static class Node {
Node left;
Node right;
int data;
Node(int newData) {
left = null;
right = null;
data = newData;
}
}
/**
Creates an empty binary tree — a null root pointer.
*/
public void BinaryTree() {
root = null;
}
/**
Returns true if the given target is in the binary tree.
Uses a recursive helper.
*/
public boolean lookup(int data) {
return(lookup(root, data));
}
/**
Recursive lookup — given a node, recur
down searching for the given data.
*/
private boolean lookup(Node node, int data) {
if (node==null) {
return(false);
}
if (data==node.data) {
return(true);
}
else if (data<node.data) {
return(lookup(node.left, data));
}
else {
return(lookup(node.right, data));
}
}
/**
Inserts the given data into the binary tree.
Uses a recursive helper.
*/
public void insert(int data) {
root = insert(root, data);
}
/**
Recursive insert — given a node pointer, recur down and
insert the given data into the tree. Returns the new
node pointer (the standard way to communicate
a changed pointer back to the caller).
*/
private Node insert(Node node, int data) {
if (node==null) {
node = new Node(data);
}
else {
if (data <= node.data) {
node.left = insert(node.left, data);
}
else {
node.right = insert(node.right, data);
}
}
return(node); // in any case, return the new pointer to the caller
}
OOP Style vs. Recursive Style
From the client point of view, the BinaryTree class demonstrates good OOP style — it encapsulates the binary tree state, and the client sends messages like lookup() and insert() to operate on that state. Internally, the Node class and the recursive methods do not demonstrate OOP style. The recursive methods like insert(Node) and lookup (Node, int) basically look like recursive functions in any language. In particular, they do not operate against a “receiver” in any special way. Instead, the recursive methods operate on the arguments that are passed in which is the classical way to write recursion. My sense is that the OOP style and the recursive style do not be combined nicely for binary trees, so I have left them separate. Merging the two styles would be especially awkward for the “empty” tree (null) case, since you can’t send a message to the null pointer. It’s possible to get around that by having a special object to represent the null tree, but that seems like a distraction to me. I prefer to keep the recursive methods simple, and use different examples to teach OOP.
Java Solutions
Here are the Java solutions to the 14 binary tree problems. Most of the solutions use two methods:a one-line OOP method that starts the computation, and a recursive method that does the real operation. Make an attempt to solve each problem before looking at the solution — it’s the best way to learn.
1. Build123() Solution (Java)
/**
Build 123 using three pointer variables.
*/
public void build123a() {
root = new Node(2);
Node lChild = new Node(1);
Node rChild = new Node(3);
root.left = lChild;
root.right= rChild;
}
/**
Build 123 using only one pointer variable.
*/
public void build123b() {
root = new Node(2);
root.left = new Node(1);
root.right = new Node(3);
}
/**
Build 123 by calling insert() three times.
Note that the ‘2’ must be inserted first.
*/
public void build123c() {
root = null;
root = insert(root, 2);
root = insert(root, 1);
root = insert(root, 3);
}
2. size() Solution (Java)
/**
Returns the number of nodes in the tree.
Uses a recursive helper that recurs
down the tree and counts the nodes.
*/
public int size() {
return(size(root));
}
private int size(Node node) {
if (node == null) return(0);
else {
return(size(node.left) + 1 + size(node.right));
}
}
3. maxDepth() Solution (Java)
/**
Returns the max root-to-leaf depth of the tree.
Uses a recursive helper that recurs down to find
the max depth.
*/
public int maxDepth() {
return(maxDepth(root));
}
private int maxDepth(Node node) {
if (node==null) {
return(0);
}
else {
int lDepth = maxDepth(node.left);
int rDepth = maxDepth(node.right);
// use the larger + 1
return(Math.max(lDepth, rDepth) + 1);
}
}
4. minValue() Solution (Java)
/**
Returns the min value in a non-empty binary search tree.
Uses a helper method that iterates to the left to find
the min value.
*/
public int minValue() {
return( minValue(root) );
}
/**
Finds the min value in a non-empty binary search tree.
*/
private int minValue(Node node) {
Node current = node;
while (current.left != null) {
current = current.left;
}
return(current.data);
}
5. printTree() Solution (Java)
/**
Prints the node values in the “inorder” order.
Uses a recursive helper to do the traversal.
*/
public void printTree() {
printTree(root);
System.out.println();
}
private void printTree(Node node) {
if (node == null) return;
// left, node itself, right
printTree(node.left);
System.out.print(node.data + “ “);
printTree(node.right);
}
6. printPostorder() Solution (Java)
/**
Prints the node values in the “postorder” order.
Uses a recursive helper to do the traversal.
*/
public void printPostorder() {
printPostorder(root);
System.out.println();
}
public void printPostorder(Node node) {
if (node == null) return;
// first recur on both subtrees
printPostorder(node.left);
printPostorder(node.right);
// then deal with the node
System.out.print(node.data + “ “);
}
7. hasPathSum() Solution (Java)
/**
Given a tree and a sum, returns true if there is a path from the root
down to a leaf, such that adding up all the values along the path
equals the given sum.
Strategy: subtract the node value from the sum when recurring down,
and check to see if the sum is 0 when you run out of tree.
*/
public boolean hasPathSum(int sum) {
return( hasPathSum(root, sum) );
}
boolean hasPathSum(Node node, int sum) {
// return true if we run out of tree and sum==0
if (node == null) {
return(sum == 0);
}
else {
// otherwise check both subtrees
int subSum = sum — node.data;
return(hasPathSum(node.left, subSum) || hasPathSum(node.right, subSum));
}
}
8. printPaths() Solution (Java)
/**
Given a binary tree, prints out all of its root-to-leaf
paths, one per line. Uses a recursive helper to do the work.
*/
public void printPaths() {
int[] path = new int[1000];
printPaths(root, path, 0);
}
/**
Recursive printPaths helper — given a node, and an array containing
the path from the root node up to but not including this node,
prints out all the root-leaf paths.
*/
private void printPaths(Node node, int[] path, int pathLen) {
if (node==null) return;
// append this node to the path array
path[pathLen] = node.data;
pathLen++;
// it’s a leaf, so print the path that led to here
if (node.left==null && node.right==null) {
printArray(path, pathLen);
}
else {
// otherwise try both subtrees
printPaths(node.left, path, pathLen);
printPaths(node.right, path, pathLen);
}
}
/**
Utility that prints ints from an array on one line.
*/
private void printArray(int[] ints, int len) {
int i;
for (i=0; i<len; i++) {
System.out.print(ints[i] + “ “);
}
System.out.println();
}
9. mirror() Solution (Java)
/**
Changes the tree into its mirror image.
So the tree…
4
/ \
2 5
/ \
1 3
is changed to…
4
/ \
5 2
/ \
3 1
Uses a recursive helper that recurs over the tree,
swapping the left/right pointers.
*/
public void mirror() {
mirror(root);
}
private void mirror(Node node) {
if (node != null) {
// do the sub-trees
mirror(node.left);
mirror(node.right);
// swap the left/right pointers
Node temp = node.left;
node.left = node.right;
node.right = temp;
}
}
10. doubleTree() Solution (Java)
/**
Changes the tree by inserting a duplicate node
on each nodes’s .left.
So the tree…
2
/ \
1 3
Is changed to…
2
/ \
2 3
/ /
1 3
/
1
Uses a recursive helper to recur over the tree
and insert the duplicates.
*/
public void doubleTree() {
doubleTree(root);
}
private void doubleTree(Node node) {
Node oldLeft;
if (node == null) return;
// do the subtrees
doubleTree(node.left);
doubleTree(node.right);
// duplicate this node to its left
oldLeft = node.left;
node.left = new Node(node.data);
node.left.left = oldLeft;
}
11. sameTree() Solution (Java)
/*
Compares the receiver to another tree to
see if they are structurally identical.
*/
public boolean sameTree(BinaryTree other) {
return( sameTree(root, other.root) );
}
/**
Recursive helper — recurs down two trees in parallel,
checking to see if they are identical.
*/
boolean sameTree(Node a, Node b) {
// 1. both empty -> true
if (a==null && b==null) return(true);
// 2. both non-empty -> compare them
else if (a!=null && b!=null) {
return(
a.data == b.data &&
sameTree(a.left, b.left) &&
sameTree(a.right, b.right)
);
}
// 3. one empty, one not -> false
else return(false);
}
12. countTrees() Solution (Java)
/**
For the key values 1…numKeys, how many structurally unique
binary search trees are possible that store those keys?
Strategy: consider that each value could be the root.
Recursively find the size of the left and right subtrees.
*/
public static int countTrees(int numKeys) {
if (numKeys <=1) {
return(1);
}
else {
// there will be one value at the root, with whatever remains
// on the left and right each forming their own subtrees.
// Iterate through all the values that could be the root…
int sum = 0;
int left, right, root;
for (root=1; root<=numKeys; root++) {
left = countTrees(root-1);
right = countTrees(numKeys — root);
// number of possible trees with this root == left*right
sum += left*right;
}
return(sum);
}
}
13. isBST1() Solution (Java)
/**
Tests if a tree meets the conditions to be a
binary search tree (BST).
*/
public boolean isBST() {
return(isBST(root));
}
/**
Recursive helper — checks if a tree is a BST
using minValue() and maxValue() (not efficient).
*/
private boolean isBST(Node node) {
if (node==null) return(true);
// do the subtrees contain values that do not
// agree with the node?
if (node.left!=null && maxValue(node.left) > node.data) return(false);
if (node.right!=null && minValue(node.right) <= node.data) return(false);
// check that the subtrees themselves are ok
return( isBST(node.left) && isBST(node.right) );
}
14. isBST2() Solution (Java)
/**
Tests if a tree meets the conditions to be a
binary search tree (BST). Uses the efficient
recursive helper.
*/
public boolean isBST2() {
return( isBST2(root, Integer.MIN_VALUE, Integer.MAX_VALUE) );
}
/**
Efficient BST helper — Given a node, and min and max values,
recurs down the tree to verify that it is a BST, and that all
its nodes are within the min..max range. Works in O(n) time —
visits each node only once.
*/
private boolean isBST2(Node node, int min, int max) {
if (node==null) {
return(true);
}
else {
// left should be in range min…node.data
boolean leftOk = isBST2(node.left, min, node.data);
// if the left is not ok, bail out
if (!leftOk) return(false);
// right should be in range node.data+1..max
boolean rightOk = isBST2(node.right, node.data+1, max);
return(rightOk);
}
}
