Balanced Binary Tree

最新推荐文章于 2023-04-13 15:01:12 发布

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最新推荐文章于 2023-04-13 15:01:12 发布

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原文链接：http://www.cnblogs.com/FLAGyuri/p/5290701.html

本文介绍了一种算法，用于判断二叉树是否为高度平衡的树。通过递归计算每个节点的左右子树深度，并确保任意节点的左右子树深度差不超过1，实现了这一功能。

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.

根据定义求解即可。需要用到binary tree depth 的算法。

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public boolean isBalanced(TreeNode root) {
12         if (root == null || root.left == null && root.right == null) {
13             return true;
14         }
15         if (Math.abs(depth(root.left) - depth(root.right)) > 1) {
16             return false;
17         }
18         return isBalanced(root.left)&&isBalanced(root.right);
19     }
20     private int depth(TreeNode root) {
21         if (root == null) {
22             return 0;
23         }
24         int result = 1;
25         return depth(root, result);
26     }
27     private int depth(TreeNode root, int result) {
28         if (root.left == null && root.right == null) {
29             return result;
30         }
31         if (root.left != null && root.right == null) {
32             return depth(root.left, result + 1);
33         } else if (root.left == null && root.right != null) {
34             return depth(root.right, result + 1);
35         } else {
36             return Math.max(depth(root.left, result + 1), depth(root.right, result + 1));
37         }
38     }
39 }