就把vector改成用栈类存放层次遍历的一层的序列

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

给定一个二叉树，自底向上的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，但是这里是从叶子节点到根节点)

例如：

给定一个二叉树 {3,9,20,#,#,15,7},

    3

   / \

  9  20

    /  \

   15   7

返回的层次遍历的结果是：

[

  [15,7]

  [9,20],

  [3],

]

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

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

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

    3

   / \

  9  20

    /  \

   15   7

return its bottom-up level order traversal as:

[

  [15,7]

  [9,20],

  [3],

]
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
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
  
#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> > levelOrderBottom(TreeNode *root)
{

vector<vector<int> > matrix;
    if(root == NULL)
    {
        return matrix;
    }

stack<vector<int> > sv;
    vector<int> temp;
    temp.push_back(root->val);
    /*先把根结点的这一行信息放到栈中*/
    sv.push(temp);

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

int count = 1;
    while(!path.empty())
    {
        if(path[0]->left != NULL)
        {
            path.push_back(path[0]->left);
        }
        if(path[0]->right != NULL)
        {
            path.push_back(path[0]->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);
            }
            sv.push(tmp);
            count = path.size();
        }
    }
    /*遍历结果栈*/
    while(!sv.empty())
    {
        if(sv.top().size() > 0)
        {
            matrix.push_back(sv.top());
        }
        sv.pop();
    }
    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 = levelOrderBottom(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：
4 7 9 2 6 1 8
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】05二叉树的层次遍历II【Binary Tree Level Order Traversal II】的更多相关文章
	

    
						
  1. [Swift]LeetCode107. 二叉树的层次遍历 II | Binary Tree Level Order Traversal II
		
    Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left ...
    
		
    2. 
						
  2. 102/107. Binary Tree Level Order Traversal/II
		
    原文题目: 102. Binary Tree Level Order Traversal 107. Binary Tree Level Order Traversal II 读题: 102. 层序遍历 ...
    
		
    3. 
						
  3. 【LeetCode】107. Binary Tree Level Order Traversal II (2 solutions)
		
    Binary Tree Level Order Traversal II Given a binary tree, return the bottom-up level order traversal ...
    
		
    4. 
						
  4. 35. Binary Tree Level Order Traversal  &&  Binary Tree Level Order Traversal II
		
    Binary Tree Level Order Traversal OJ: https://oj.leetcode.com/problems/binary-tree-level-order-trave ...
    
		
    5. 
						
  5. Binary Tree Level Order Traversal,Binary Tree Level Order Traversal II
		
    Binary Tree Level Order Traversal Total Accepted: 79463 Total Submissions: 259292 Difficulty: Easy G ...
    
		
    6. 
						
  6. LeetCode之“树”：Binary Tree Level Order Traversal && Binary Tree Level Order Traversal II
		
    Binary Tree Level Order Traversal 题目链接 题目要求: Given a binary tree, return the level order traversal o ...
    
		
    7. 
						
  7. LeetCode_107. Binary Tree Level Order Traversal II
		
    107. Binary Tree Level Order Traversal II Easy Given a binary tree, return the bottom-up level order ...
    
		
    8. 
						
  8. 63. Binary Tree Level Order Traversal II
		
    Binary Tree Level Order Traversal II My Submissions QuestionEditorial Solution Total Accepted: 79742 ...
    
		
    9. 
						
  9. [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 ...
    
		
    10. 
						
  10. [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 ...
    
		
    11. 
		

	


随机推荐
	

    
									
  1. php 前台post多维数组
			
    post多维数组时,在后台用$_POST接收不正常,应使用$_REQUEST
    
			
    2. 
						
  2. 做完task1-21的阶段总结
			
    [说明]这是自注册修真院的第七天,也是第七篇日报,觉得是一个好的时机总结一下. 因为任务一虽然看起来仅仅是“完成学员报名的DB设计并读写数据库”,但是做了几天之后就发现在任务“搭建自己的服务器”之前的 ...
    
			
    3. 
						
  3. Jmeter+Ant+Jenkins接口自动化测试框架搭建
			
    前言 软件开发的V模型大家都不陌生,其中测试阶段分为单元测试→功能测试→系统测试→验收测试.其中单元测试一般由开发同学们自己完成,大部分测试具体实施(这里不包括用例设计)是从单体功能测试开始着手的.  ...
    
			
    4. 
						
  4. [CTSC2001]1378 选课
			
      1378 选课 题目描述 学校实行学分制.每门的必修课都有固定的学分,同时还必须获得相应的选修课程学分.学校开设了N(N<300)门的选修课程,每个学生可选课程的数量M是给定的.学生选修了这 ...
    
			
    5. 
						
  5. PAT 1063. 计算谱半径(20)
			
    在数学中,矩阵的“谱半径”是指其特征值的模集合的上确界.换言之,对于给定的n个复数空间的特征值{a1+b1i, ..., an+bni},它们的模为实部与虚部的平方和的开方,而“谱半径”就是最大模.  ...
    
			
    6. 
						
  6. linux c编程：进程控制(四)进程关系
			
    每一个进程除了有一个进程ID外,还属于一个进程组.  进程组是一个或多个进程的集合,通常情况下,他们是在同一作业中结合起来的,同一进程组的个进程接受来自同一终端的各种信号. 每一个进程组有一个唯一的进 ...
    
			
    7. 
						
  7. ABAP 面向对象(Object Orientation) OO
			
    [转自 http://blog.sina.com.cn/s/blog_7c7b16000101bhof.html]在程序中, 对象的识别和寻址是通过对象引用来实现的, 对象引用变量可以访问对象的属性和 ...
    
			
    8. 
						
  8. php正则表达式和数组
			
    一.正则表达式 1. “/”代表定界符,"^"代表起始符号,"$"代表结束符号 $str1="abc123def45ghjk6789lou" ...
    
			
    9. 
						
  9. 【leetcode刷题笔记】Word Search
			
    Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from l ...
    
			
    10. 
						
  10. MySQL 及 SQL 注入与防范方法
			
    所谓SQL注入,就是通过把SQL命令插入到Web表单递交或输入域名或页面请求的查询字符串,最终达到欺骗服务器执行恶意的SQL命令. 我们永远不要信任用户的输入,我们必须认定用户输入的数据都是不安全的, ...
    
			
    11.