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

给定一个二叉树，返回他的层次遍历的节点的values。(提示，从左到右，一层一层的遍历)

例如：

给定一个二叉树 {1,#,2,3},

   1

    \

     2

    /

   3

返回的层次遍历的结果是：

[

  [3],

  [9,20],

  [15,7]

]

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

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3

   / \

  9  20

    /  \

   15   7

return its level order traversal as:

[

  [3],

  [9,20],

  [15,7]

]
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

【二叉树遍历模版】层次遍历

count 记录的是当前遍历的层次当中的结点个数。

depth记录的是当前遍历过的层次数。

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
  
#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) {}
 * };
 */
vector<vector<int> > levelOrder(TreeNode *root)
{

vector<vector<int> > matrix;
    if(root == NULL)
    {
        return matrix;
    }
    vector<int> temp;
    temp.push_back(root->val);
    matrix.push_back(temp);

vector<TreeNode *> path;
    path.push_back(root);

int count = 1;
    while(!path.empty())
    {
        TreeNode *tn = path.front();
        if(tn->left)
        {
            path.push_back(tn->left);
        }
        if(tn->right)
        {
            path.push_back(tn->right);
        }
        path.erase(path.begin());
        count--;

if(count == 0)
        {
            vector<int> tmp;
            vector<TreeNode *>::iterator it = path.begin();
            for(; it != path.end(); ++it)
            {
                tmp.push_back((*it)->val);
            }
            if(tmp.size() > 0)
            {
                matrix.push_back(tmp);
            }
            count = path.size();
        }
    }
    return matrix;
}

// 树中结点含有分叉，
//                  8
//              /       \
//             6         1
//           /   \
//          9     2
//               / \
//              4   7
int main()
{
    TreeNode *pNodeA1 = CreateBinaryTreeNode(8);
    TreeNode *pNodeA2 = CreateBinaryTreeNode(6);
    TreeNode *pNodeA3 = CreateBinaryTreeNode(1);
    TreeNode *pNodeA4 = CreateBinaryTreeNode(9);
    TreeNode *pNodeA5 = CreateBinaryTreeNode(2);
    TreeNode *pNodeA6 = CreateBinaryTreeNode(4);
    TreeNode *pNodeA7 = CreateBinaryTreeNode(7);

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

PrintTree(pNodeA1);

vector<vector<int> > ans = levelOrder(pNodeA1);

for (int i = 0; i < ans.size(); ++i)
    {
        for (int j = 0; j < ans[i].size(); ++j)
        {
            cout << ans[i][j] << " ";
        }
    }
    cout << endl;

DestroyTree(pNodeA1);
    return 0;
}
输出结果：
8 6 1 9 2 4 7
 
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】04二叉树的层次遍历【Binary Tree Level Order Traversal】的更多相关文章
	

    
						
  1. [LintCode] Binary Tree Level Order Traversal（二叉树的层次遍历）
		
    描述 给出一棵二叉树,返回其节点值的层次遍历(逐层从左往右访问) 样例 给一棵二叉树 {3,9,20,#,#,15,7} : 3 / \ 9 20 / \ 15 7 返回他的分层遍历结果: [ [3] ...
    
		
    2. 
						
  2. LeetCode 102. 二叉树的层次遍历(Binary Tree Level Order Traversal) 8
		
    102. 二叉树的层次遍历 102. Binary Tree Level Order Traversal 题目描述 给定一个二叉树,返回其按层次遍历的节点值. (即逐层地,从左到右访问所有节点). 每 ...
    
		
    3. 
						
  3. [Leetcode] Binary tree level order traversal ii二叉树层次遍历
		
    Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left ...
    
		
    4. 
						
  4. [LeetCode] Binary Tree Level Order Traversal II 二叉树层序遍历之二
		
    Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left ...
    
		
    5. 
						
  5. [LeetCode] Binary Tree Level Order Traversal 二叉树层序遍历
		
    Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, ...
    
		
    6. 
						
  6. LeetCode之Binary Tree Level Order Traversal  层序遍历二叉树
		
    Binary Tree Level Order Traversal 题目描述: Given a binary tree, return the level order traversal of its ...
    
		
    7. 
						
  7. LeetCode 107 Binary Tree Level Order Traversal II（二叉树的层级顺序遍历2）（*）
		
    翻译 给定一个二叉树,返回从下往上遍历经过的每一个节点的值. 从左往右,从叶子到节点. 比如: 给定的二叉树是 {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 返回它从下 ...
    
		
    8. 
						
  8. [LeetCode] 102. Binary Tree Level Order Traversal 二叉树层序遍历
		
    Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, ...
    
		
    9. 
						
  9. [LeetCode] 107. Binary Tree Level Order Traversal II 二叉树层序遍历 II
		
    Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left ...
    
		
    10. 
						
  10. [LeetCode] Binary Tree Level Order Traversal 与 Binary Tree Zigzag Level Order Traversal，两种按层次遍历树的方式，分别两个队列，两个栈实现
		
    Binary Tree Level Order Traversal Given a binary tree, return the level order traversal of its nodes ...
    
		
    11. 
		

	


随机推荐
	

    
									
  1. UGUI随记
			
    <color=#ffef00ff>武器</color>:巨剑 <color=#ffef00ff>种族</color>:人族 <color=#ffe ...
    
			
    2. 
						
  2. Lumen开发：phpunit单元测试
			
    先来直接运行,cmd先进入根目录,然后进入tests或是test文件夹 运行命令行:..\vendor\bin\phpunit ExampleTest.php laravel/lumen中集成了PHP ...
    
			
    3. 
						
  3. EasyDSS视频点播服务器实现多分辨率/多码率无缝切换的办法
			
    EasyDSS流媒体音视频直播与点播服务器软件,是一套提供一站式的转码.点播.直播.检索.回放.录像下载服务的高性能RTMP/HLS/HTTP-FLV流媒体服务,极大地简化了流媒体相关业务的开发和集成 ...
    
			
    4. 
						
  4. python基础-第五篇-5.1冒泡排序
			
    几个月过去了,小白逐渐对公司的后端服务熟悉了,不过这天小白又接到一封神秘邮件,是景女神发来的:公司急需一批对语言算法有些了解的优秀员工,鉴于你在公司的表现很不错,现在给到你一个培训机会,请速到开发部报 ...
    
			
    5. 
						
  5. R语言编写乘法表
			
    for(i in 1:9){ for(j in 1:i){ m = j*i cat(i,'*',j,'=',m,' ') } cat('\n') } 1 * 1 = 1 2 * 1 = 2 2 * 2 ...
    
			
    6. 
						
  6. QT5的QDesktopSerivices不同
			
    QT4使用QDesktopServices::storageLocation(QDesktopServices::xxxx)来获取一些系统目录, 现在则要改成QStandardPaths::writa ...
    
			
    7. 
						
  7. ubuntu sudo-update出错Encountered a section with no Package: header
			
    Reading package lists... Error! E: Encountered a section with no Package: header E: Problem with Mer ...
    
			
    8. 
						
  8. 【LeetCode】【找元素】Find First and Last Position of Element in Sorted Array
			
    描述: Given an array of integers nums sorted in ascending order, find the starting and ending position ...
    
			
    9. 
						
  9. Data Structure Graph: cycle in a directed graph
			
    geeks上的解答复杂了些,用回溯就行了 #include <iostream> #include <vector> #include <algorithm> #i ...
    
			
    10. 
						
  10. [原创]java WEB学习笔记25：MVC案例完整实践(part 6)---新增操作的设计与实现
			
    本博客为原创:综合 尚硅谷(http://www.atguigu.com)的系统教程(深表感谢)和 网络上的现有资源(博客,文档,图书等),资源的出处我会标明 本博客的目的:①总结自己的学习过程,相当 ...
    
			
    11.