Binary Tree Inorder Traversal

Problem Statement

Given a binary tree, return the inorder traversal of its nodes’ values.

For example:
Given binary tree {1,#,2,3},

1
    \
     2
    /
   3

return [1,3,2].

Approach

There are 2 ways I know of solving this problem:

• Maintain of stack of the left most nodes.
• Pop the top element from the stack-> add the value of this node to the list -> push the right subtree of this node into the stack.
• Perform the above step until the stack is empty. The elements added to the list will be the result of inorder traversal.

Another approach is simple:

• Starting from the root node, keep adding to the list in the following order
• If left node is not null then recursively call addToList(node.left)
• Add value of the current node to the list.
• If right node is not null then recursively call addToList(node.right)

In this approach there is no need of maintaining a stack.

Run Time Complexity

The run time for the first approach is O(n logn). Cost of pushToLeft() is O(lgn) and this called for n elements so the totals becomes O(nlogn)

The run time for the second approach is clearly O(n)

Code for Approach1

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    Stack<TreeNode> stack = new Stack<TreeNode>();
    List<Integer> result = new ArrayList<Integer>();
    
    public List<Integer> inorderTraversal(TreeNode root) {
        if(root == null){
            return result;
        }
        pushToLeft(root);
        while(!stack.empty()){
            TreeNode node = stack.pop();
            result.add(node.val);
            pushToLeft(node.right);
        }
        return result;
    }
    
    private void pushToLeft(TreeNode node){
        if(node != null){
            stack.push(node);
            pushToLeft(node.left);
        }return;
    }}

Code for Approach2

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    
    
    List<Integer> result = new ArrayList<Integer>();
    public List<Integer> inorderTraversal(TreeNode root) {
        if(root == null){
            return result;
        }
        addToList(root);
        return result;
    }
    
    private void addToList(TreeNode node){
        if(node.left != null){
            addToList(node.left);
        }
        result.add(node.val);
        
        if(node.right != null){
            addToList(node.right);
        }
    }
}