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/beiyeqingteng/p/5652091.html

本文介绍了一种算法，用于判断一棵二叉树是否为高度平衡的二叉树。通过递归计算每个节点的左右子树的高度差，确保任意节点的左右子树高度差不超过1。提供了两种实现方式，一种直接利用高度函数进行判断，另一种采用深度优先搜索优化。

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.

Example

Given binary tree A = {3,9,20,#,#,15,7}, B = {3,#,20,15,7}

A)  3            B)    3 
   / \                  \
  9  20                 20
    /  \                / \
   15   7              15  7

The binary tree A is a height-balanced binary tree, but B is not.

 1 /**
 2  * Definition of TreeNode:
 3  * public class TreeNode {
 4  *     public int val;
 5  *     public TreeNode left, right;
 6  *     public TreeNode(int val) {
 7  *         this.val = val;
 8  *         this.left = this.right = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     /**
14      * @param root: The root of binary tree.
15      * @return: True if this Binary tree is Balanced, or false.
16      */
17     public boolean isBalanced(TreeNode root) {  
18         if (root == null) return true;  
19         if (Math.abs(height(root.left) - height(root.right)) > 1) return false;  
20         return isBalanced(root.left) && isBalanced(root.right);  
21     }  
22       
23     public int height(TreeNode root) {  
24         if (root == null) return 0;  
25         return Math.max(height(root.left), height(root.right)) + 1;  
26     }  
27 }

 1 public class Solution {
 2     public boolean isBalanced(TreeNode root) {
 3         if (root == null) return true;
 4         return helper(root) != -1;
 5     }
 6     
 7     public int helper(TreeNode root) {
 8         if (root == null) return 0;
 9         int leftHeight = helper(root.left);
10         int rightHeight = helper(root.right);
11         
12         if (leftHeight == -1 || rightHeight == -1) return -1;
13         if (Math.abs(leftHeight - rightHeight) > 1) return -1;
14         
15         return Math.max(helper(root.left), helper(root.right)) + 1;
16     }
17 }