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

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. 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. [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. [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. [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. 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. 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. [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. [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. [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 ...

随机推荐

  1. unity 常用的几种相机跟随

    固定相机跟随 这种相机有一个参考对象,它会保持与该参考对象固定的位置,跟随改参考对象发生移动 using UnityEngine; using System.Collections; public c ...

  2. Java是否存在内存泄露

    会的. 原因:长生命周期的对象持有短生命周期对象的引用,导致短生命周期对象不能被回收,由此可能发生内存泄露. 举例参考:http://blog.csdn.net/yakihappy/article/d ...

  3. 让Xcode支持高版本系统设备真机测试

    最新支持11.2 (15C107) Xcode只可以支持iPhone手机对应iOS系统以下的真机测试.一般想要支持最新的iPhone手机系统,有两个方法. 第一.就需要更新Xcode,这一个方法有一个 ...

  4. [JavaScript]WebBrowser控件下IE版本的检测

    转载请注明原文地址:https://www.cnblogs.com/litou/p/10772272.htm 在客户端检查用户使用的浏览器类型和版本,都是根据navigator.userAgent属性 ...

  5. php扩展trie_filter: 利用词库, 过滤敏感词

    1. 先安装libiconv# wget http://ftp.gnu.org/pub/gnu/libiconv/libiconv-1.13.1.tar.gz# tar -zxvf libiconv- ...

  6. activiti踩坑

    最近在学习activiti,偶然间遇到一个错误:加载引擎的时候报错,显示空指针错误,跟代码发现初始化配置文件返回为null.几经排查,可能是因为我发布流程后又清空了数据库数据导致的.然后我把表全部删除 ...

  7. 第13条:合理利用try/expect/else/finally结构中的每个代码块

    核心知识点: (1)无论try块是否发生异常,都可以使用try/finally复合语句中地finally块来执行清理工作. (2)顺利运行try块后,若想使某些操作能在finally块地清理代码之前执 ...

  8. monokai-background

    foreground-color:f8f8f2 background-color:272822

  9. 数据库 简单查询 Sql Server 学生表 课程表 选课表

    创建教材中的三张表格,并输入相应的数据 Create table student( Sno char(9), Same char(20), Ssex char(2), Sage smallint, S ...

  10. DEV开发之控件NavBarControl

    右键点击RunDesigner弹出如下界面鼠标先点击3或4,1,,然后点击1或2进行相应的新增或删除操作,3是分组,4是项目,4可以直接拖动到相应的分组3.属性caption:显示的名称4.NavBa ...