110. Balanced Binary Tree
Published in
2 min readJul 19, 2023
Easy
Given a binary tree, determine if it is height-balanced.
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: true
Example 2:
Input: root = [1,2,2,3,3,null,null,4,4]
Output: false
Example 3:
Input: root = []
Output: true
Constraints:
- The number of nodes in the tree is in the range
[0, 5000]
. -104 <= Node.val <= 104
Solution:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
// Get the height of left and right subtrees using dfs traversal. Return true if the // difference between heights is not more than 1 and left and right subtrees are
// balanced, otherwise return false.
class Solution {
public boolean isBalanced(TreeNode root) {
if(root == null)
return true;
int l = height(root.left);
int r = height(root.right);
if(Math.abs(l-r) <=1 && isBalanced(root.left) && isBalanced(root.right))
return true;
else
return false;
}
public int height(TreeNode root){
if(root == null)
return 0;
else
return 1+ Math.max(height(root.left),height(root.right));
}
}