110. 平衡二叉树
Difficulty: 简单
给定一个二叉树，判断它是否是高度平衡的二叉树。
本题中，一棵高度平衡二叉树定义为：
一个二叉树_每个节点 _的左右两个子树的高度差的绝对值不超过 1 。
示例 1：
输入：root = [3,9,20,null,null,15,7]
输出：true
示例 2：
输入：root = [1,2,2,3,3,null,null,4,4]
输出：false
示例 3：
输入：root = []
输出：true
提示：
- 树中的节点数在范围
[0, 5000]内 -10<sup>4</sup> <= Node.val <= 10<sup>4</sup>
Solution
Language: java
/**
* 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 postOrder(root).isBalanced;
}
private Twotuple postOrder(TreeNode root){
if(root == null) return new Twotuple(0, true);
Twotuple left = postOrder(root.left);
if(!left.isBalanced) return new Twotuple(0, false);
Twotuple right = postOrder(root.right);
if(!right.isBalanced) return new Twotuple(0, false);
if(Math.abs(left.heigh - right.heigh) > 1) return new Twotuple(0, false);
return new Twotuple(Math.max(left.heigh, right.heigh) + 1, true);
}
class Twotuple{
int heigh;
boolean isBalanced;
public Twotuple(int h, boolean b){
this.heigh = h;
this.isBalanced = b;
}
}
}