平衡的二叉树的定义都是递归的定义,所以,用递归来解决问题,还是挺容易的额。

本质上是递归的遍历二叉树。

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

给定一个二叉树,判定他是不是高度平衡的二叉树。

对于这个问题,每个节点的两个子树的深度不会相差超过1,那么这样的二叉树就是一个平衡的二叉树

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

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 depth of the two subtrees of every node never differ by more than 1.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
test.cpp:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
 
#include <iostream>
#include <cstdio>
#include <stack>
#include <vector>
#include "BinaryTree.h"

using namespace std;

/**
 * Definition for binary tree
 * struct TreeNode {
 * int val;
 * TreeNode *left;
 * TreeNode *right;
 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
bool balance(TreeNode *root, int &depth)
{
    if(root == NULL)
    {
        depth = 0;
        return true;
    }
    int leftdepth = 0;
    bool left = balance(root->left, leftdepth);

int rightdepth = 0;
    bool right = balance(root->right, rightdepth);

depth = leftdepth > rightdepth ? leftdepth + 1 : rightdepth + 1;
    int gap = leftdepth - rightdepth;

return left && right && (-1 <= gap && gap <= 1);
}

bool isBalanced(TreeNode *root)
{

int depth = 0;
    return balance(root, depth);
}

// 树中结点含有分叉,
//                  6
//              /       \
//             7         2
//           /   \
//          1     4
//               / \
//              3   5
int main()
{
    TreeNode *pNodeA1 = CreateBinaryTreeNode(6);
    TreeNode *pNodeA2 = CreateBinaryTreeNode(7);
    TreeNode *pNodeA3 = CreateBinaryTreeNode(2);
    TreeNode *pNodeA4 = CreateBinaryTreeNode(1);
    TreeNode *pNodeA5 = CreateBinaryTreeNode(4);
    TreeNode *pNodeA6 = CreateBinaryTreeNode(3);
    TreeNode *pNodeA7 = CreateBinaryTreeNode(5);

ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3);
    ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5);
    ConnectTreeNodes(pNodeA5, pNodeA6, pNodeA7);

// 树中结点含有分叉,
    //                  1
    //              /       \
    //             2         2
    //           /   \       / \
    //          3     4     4   3

TreeNode *pNodeB1 = CreateBinaryTreeNode(1);
    TreeNode *pNodeB2 = CreateBinaryTreeNode(2);
    TreeNode *pNodeB3 = CreateBinaryTreeNode(2);
    TreeNode *pNodeB4 = CreateBinaryTreeNode(3);
    TreeNode *pNodeB5 = CreateBinaryTreeNode(4);
    TreeNode *pNodeB6 = CreateBinaryTreeNode(4);
    TreeNode *pNodeB7 = CreateBinaryTreeNode(3);

ConnectTreeNodes(pNodeB1, pNodeB2, pNodeB3);
    ConnectTreeNodes(pNodeB2, pNodeB4, pNodeB5);
    ConnectTreeNodes(pNodeB3, pNodeB6, pNodeB7);

bool ans = isBalanced(pNodeA1);

if (ans == true)
    {
        cout << "Balanced!" << endl;
    }
    else
    {
        cout << "Not Balanced!" << endl;
    }

bool ans1 = isBalanced(pNodeB1);

if (ans1 == true)
    {
        cout << "Balanced!" << endl;
    }
    else
    {
        cout << "Not Balanced!" << endl;
    }
    DestroyTree(pNodeA1);
    DestroyTree(pNodeB1);
    return 0;
}

结果输出:
Not Balanced!
Balanced!
 
BinaryTree.h:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
 
#ifndef _BINARY_TREE_H_
#define _BINARY_TREE_H_

struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

TreeNode *CreateBinaryTreeNode(int value);
void ConnectTreeNodes(TreeNode *pParent,
                      TreeNode *pLeft, TreeNode *pRight);
void PrintTreeNode(TreeNode *pNode);
void PrintTree(TreeNode *pRoot);
void DestroyTree(TreeNode *pRoot);

#endif /*_BINARY_TREE_H_*/

BinaryTree.cpp:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
 
#include <iostream>
#include <cstdio>
#include "BinaryTree.h"

using namespace std;

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

//创建结点
TreeNode *CreateBinaryTreeNode(int value)
{
    TreeNode *pNode = new TreeNode(value);

return pNode;
}

//连接结点
void ConnectTreeNodes(TreeNode *pParent, TreeNode *pLeft, TreeNode *pRight)
{
    if(pParent != NULL)
    {
        pParent->left = pLeft;
        pParent->right = pRight;
    }
}

//打印节点内容以及左右子结点内容
void PrintTreeNode(TreeNode *pNode)
{
    if(pNode != NULL)
    {
        printf("value of this node is: %d\n", pNode->val);

if(pNode->left != NULL)
            printf("value of its left child is: %d.\n", pNode->left->val);
        else
            printf("left child is null.\n");

if(pNode->right != NULL)
            printf("value of its right child is: %d.\n", pNode->right->val);
        else
            printf("right child is null.\n");
    }
    else
    {
        printf("this node is null.\n");
    }

printf("\n");
}

//前序遍历递归方法打印结点内容
void PrintTree(TreeNode *pRoot)
{
    PrintTreeNode(pRoot);

if(pRoot != NULL)
    {
        if(pRoot->left != NULL)
            PrintTree(pRoot->left);

if(pRoot->right != NULL)
            PrintTree(pRoot->right);
    }
}

void DestroyTree(TreeNode *pRoot)
{
    if(pRoot != NULL)
    {
        TreeNode *pLeft = pRoot->left;
        TreeNode *pRight = pRoot->right;

delete pRoot;
        pRoot = NULL;

DestroyTree(pLeft);
        DestroyTree(pRight);
    }
}

 
 
 

 
 

【遍历二叉树】10判断二叉树是否平衡【Balanced Binary Tree】的更多相关文章

  1. C++版 - 剑指offer 面试题39:判断平衡二叉树(LeetCode 110. Balanced Binary Tree) 题解

    剑指offer 面试题39:判断平衡二叉树 提交网址:  http://www.nowcoder.com/practice/8b3b95850edb4115918ecebdf1b4d222?tpId= ...

  2. [LeetCode] 110. Balanced Binary Tree ☆(二叉树是否平衡)

    Balanced Binary Tree [数据结构和算法]全面剖析树的各类遍历方法 描述 解析 递归分别判断每个节点的左右子树 该题是Easy的原因是该题可以很容易的想到时间复杂度为O(n^2)的方 ...

  3. 110.Balanced Binary Tree Leetcode解题笔记

    110.Balanced Binary Tree Given a binary tree, determine if it is height-balanced. For this problem, ...

  4. 平衡二叉树(Balanced Binary Tree 或 Height-Balanced Tree)又称AVL树

    平衡二叉树(Balanced Binary Tree 或 Height-Balanced Tree)又称AVL树 (a)和(b)都是排序二叉树,但是查找(b)的93节点就需要查找6次,查找(a)的93 ...

  5. [LeetCode] 110. Balanced Binary Tree 平衡二叉树

    Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary ...

  6. LeetCode 110. 平衡二叉树(Balanced Binary Tree) 15

    110. 平衡二叉树 110. Balanced Binary Tree 题目描述 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 ...

  7. 110. Balanced Binary Tree - LeetCode

    Question 110. Balanced Binary Tree Solution 题目大意:判断一个二叉树是不是平衡二叉树 思路:定义个boolean来记录每个子节点是否平衡 Java实现: p ...

  8. [LeetCode] 110. Balanced Binary Tree 解题思路

    Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary ...

  9. 【leetcode】Balanced Binary Tree(middle)

    Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary ...

  10. 33. Minimum Depth of Binary Tree && Balanced Binary Tree && Maximum Depth of Binary Tree

    Minimum Depth of Binary Tree OJ: https://oj.leetcode.com/problems/minimum-depth-of-binary-tree/ Give ...

随机推荐

  1. ASP.NET动态网站制作(19)-- C#(2)

    前言:C#的第二次课,依旧讲解C#的基础知识. 内容: 1.GC:垃圾回收机制,可以回收托管模块中的垃圾. 2.值类型和引用类型:  (1)值类型:所有的数值类型都是值类型,如int,byte,sho ...

  2. vue2 本地安装

  3. 走进EC6的let和const命令

    详细学习链接: http://es6.ruanyifeng.com/#docs/let let命令 基本用法 ES6新增了let命令,用来声明变量.它的用法类似于var,但是所声明的变量,只在let命 ...

  4. Leetcode - CopyWithRandomList

    Algorithm: Iterate and copy the original list first. For the random pointer, just copy the value fro ...

  5. centos7.0安装redis扩展

    1.下载 下载地址:https://github.com/phpredis/phpredis/ 文件名:phpredis-develop.zip 文件下载成功后,上传至/usr/local 2.安装 ...

  6. centos7.0 安装nginx

    在centos7.0下安装nginx需要安装 prce和zlib包去官网下载相应的包 然后解压相应的包进行编译 解压nginx源码包进入到解压文件 ./configure --sbin-path=/u ...

  7. 【BZOJ2401】陶陶的难题I 欧拉函数+线性筛

    [BZOJ2401]陶陶的难题I 题意:求,n<=1000000,T<=100000 题解:直接做是n*sqrt(n)的,显然会TLE,不过这题a和b都是循环到n,那么就可以进行如下的神奇 ...

  8. cocos2d-x 源代码分析 : control 源代码分析 ( 控制类组件 controlButton)

    源代码版本号来自3.1rc 转载请注明 cocos2d-x源代码分析总文件夹 http://blog.csdn.net/u011225840/article/details/31743129 1.继承 ...

  9. Swift学习 --- 2.1基础部分

    1.swift 能够省去; 2.println与print的差别就是一个能够换行一个不能够 3.swift省去了.h与.m 直接一个swift文件 4.元组能够返回多个值,元组(tuples)把多个值 ...

  10. python发布包流程

    1.新建文件夹suba和subb,文件夹下新建__init__.py,内容可以为空 2.suba内新建文件aa.py bb.py 3.subb内新建文件cc.py dd.py 4.setup.py文件 ...