110. Balanced Binary Tree 110.平衡二叉树
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 left and right subtrees of every node differ in height by no more than 1.
Example 1:
Given the following tree [3,9,20,null,null,15,7]
:
3
/ \
9 20
/ \
15 7
Return true.
Example 2:
Given the following tree [1,2,2,3,3,null,null,4,4]
:
1 / \ 2 2 / \ 3 3 / \ 4 4
复习时还不会的地方:
为啥是max + 1呢,我觉得有时候是left + 1,有时候是right + 1
这个helper是每次+1的意思吗?是的吧,每次+1并且只+1,直到尽头。那肯定只有二者间较长的可以加了
一开始以为是最长路径和最短路径相减,不懂为啥直接是左右的最长路径相减了:
不是,只要任意一个|left - right| > 1就不算。所以随时准备返回false
然后还是用求二叉树最大长度的那个方法
有个奇葩的边界情况:左或者右节点为空时,长度返回-1
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
/**
* 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) {
//cc
if (root == null) return true;
return (helper(root) != -1);
}
public int helper(TreeNode root) {
//cc
if (root == null) return 0;
int left = helper(root.left);
int right = helper(root.right);
if (left == -1 || right == -1 || Math.abs(left - right) > 1)
return -1;
return Math.max(left, right) + 1;
}
}