110. 平衡二叉树
查看题目简单
树
二叉树
低频
解法一：自底向上+递归法
时间复杂度：O(n) | 空间复杂度：O(h) | 推荐使用
动画演示
准备就绪 - 输入二叉树的JSON格式，然后点击开始
代码实现
/**
* 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;
* }
* }
*/
class Solution {
public boolean isBalanced(TreeNode root) {
return height(root) >= 0;
}
public int height(TreeNode root) {
if (root == null) {
return 0;
}
int leftHeight = height(root.left);
int rightHeight = height(root.right);
if (leftHeight == -1 || rightHeight == -1 || Math.abs(leftHeight - rightHeight) > 1) {
return -1;
} else {
return Math.max(leftHeight, rightHeight) + 1;
}
}
}
时间复杂度：O(n)
空间复杂度：O(h)