[LeetCode] 110. 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 depth of the two subtrees of every node never differ by more than 1.

Thought process:
Divide and conquer. We can build the solution on top of "104. Maximum Depth of Binary Tree". The idea is to use a function called getDepth, which returns the depth of a tree node. We then compare the depths of a node's left subtree and right subtree. There are two cases:
  1. Tree is balanced. Return maximum depth of the node.
  2. Otherwise, return -1 to signal calling function that tree is not balanced.

Solution:

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

Time complexity: O(n).
Location: Philadelphia, PA, USA

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