leetcode - Balanced Binary Tree

题目：Balanced Binary Tree

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.

 

个人思路：

1、判断每个节点（子树）的高度差，高度差在绝对值为1的范围内便是平衡二叉树

2、可以适当改造计算树高度的方法，即树的高度为左子树与右子树高度较大者加1

 

代码：

 1 #include <stddef.h>
 2 #include <iostream>
 3 /**
 4  * Definition for binary tree
 5  * struct TreeNode {
 6  *     int val;
 7  *     TreeNode *left;
 8  *     TreeNode *right;
 9  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
10  * };
11  */
12 
13 struct TreeNode
14 {
15     int val;
16     TreeNode *left;
17     TreeNode *right;
18     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
19 };
20 
21 class Solution
22 {
23 public:
24     bool isBalanced(TreeNode *root)
25     {
26         balanced = true;
27         getDepth(root);
28 
29         return balanced;
30     }
31     int getDepth(TreeNode *root)
32     {
33         if (root == NULL)
34         {
35             return 0;
36         }
37 
38         int leftDepth = getDepth(root->left);
39         int rightDepth = getDepth(root->right);
40 
41         if (leftDepth - rightDepth > 1 || leftDepth - rightDepth < -1)
42         {
43             balanced = false;
44         }
45 
46         return leftDepth > rightDepth ? leftDepth + 1 : rightDepth + 1;
47     }
48 private:
49     bool balanced;
50 };
51 
52 int main()
53 {
54     TreeNode *root = new TreeNode(1);
55     root->right = new TreeNode(1);
56     root->right->right = new TreeNode(1);
57     Solution s;
58     s.isBalanced(root);
59     std::cout << root->val << std::endl << root->right->val << std::endl << root->right->val << std::endl;
60     system("pause");
61 
62     return 0;
63 }
View Code

 

 上网搜了一些帖子，方法都是类似，就不贴出来了

posted on 2014-06-18 21:36  laihaiteng  阅读(169)  评论(0)  编辑  收藏  举报

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