Leetcode: Balanced Binary Tree

最新推荐文章于 2025-08-09 17:01:50 发布

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本文介绍了如何通过递归和改进的方法判断一个二叉树是否为高度平衡的二叉树，详细解释了高度平衡二叉树的定义，并提供了一种更优的实现策略。

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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.

Below solution uses two recursion, which is not recommended: find the max depth of left and right node separately, and compare to see their difference. Actually recursion is repeated here, which can be improved.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isBalanced(TreeNode root) {
        if (root == null) {
            return true;
        }
        
        int left = maxDepth(root.left);
        int right = maxDepth(root.right);
        if (Math.abs(left - right) > 1) {
            return false;
        } else {
            return isBalanced(root.left) && isBalanced(root.right);
        }
        
    }
    
    private int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
    }
}

A preferable method is: when calculating for the max depth, exam whether the difference of depth of left subtree and right subtree is larger than 1 at the same time. Once it is, return -1 as a false answer.

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