LeetCode 110. Balanced Binary Tree

Description

https://leetcode.com/problems/balanced-binary-tree/

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as:

a binary tree in which the left and right subtrees of every node differ in height by no more than 1.

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

Explanation

Check if left and right subtrees are balanced and whether their height difference is no more than 1.

Java Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        return maxDepth(root) != null;        
    }
    
    private Integer maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        Integer leftDepth = maxDepth(root.left);
        Integer rightDepth = maxDepth(root.right);
        
        if (leftDepth == null || rightDepth == null) {
            return null;
        }
        
        if (Math.abs(leftDepth - rightDepth) > 1) {
            return null;
        }
        
        return Math.max(leftDepth, rightDepth) + 1;
    }
}

Python Solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isBalanced(self, root: TreeNode) -> bool:

        if not root:
            return True
        
        if not root.left and not root.right:
            return True
                
        left_height = self.height_helper(root.left)
        right_height = self.height_helper(root.right)
        
        
        return self.isBalanced(root.left) and self.isBalanced(root.right) and abs(left_height - right_height) <= 1
        

        
    def height_helper(self, root):
        if not root:
            return 0
        
        if not root.left and not root.right:
            return 1
        
        
        left = self.height_helper(root.left)
        
        right = self.height_helper(root.right)
        
        return max(left, right) + 1
        

• Time Complexity: O(N)
• Space Complexity: O(N)