[leetcode] Balanced Binary Tree

原创于 2013-04-28 14:08:57 发布 · 132 阅读

0 ·

CC 4.0 BY-SA版权

java 同时被 2 个专栏收录

80 篇文章

订阅专栏

algorithm

34 篇文章

订阅专栏

本文介绍了一种高效的算法来检查二叉树是否平衡。该算法通过递归地计算每个节点的高度，并比较左右子树的高度差来实现。当左右子树高度差大于1时，则认为该二叉树不平衡。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Check if a binary tree is balanced or not. This solution is from the discuss board. Much better than mine.


/**
 * 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 height(root) != -1;
    }

    private int height(TreeNode root)
    {
        if(root == null)
            return 0;

        int leftHeight = height(root.left);
        if(leftHeight == -1)
            return -1;

        int rightHeight = height(root.right);
        if(rightHeight == -1)
            return -1;

        if(Math.abs(leftHeight - rightHeight) > 1)
            return -1;

        return 1 + Math.max(leftHeight, rightHeight);
    }
}