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. textarea中的内容的获取
			
    今天他们说为啥获取不到textarea的数值 这个问题让我很纳闷  为什么会获取不到呢? 按照逻辑来说 同样都是表单元素  怎么可能出现呢? 我就看了一眼代码 alert($("#texta ...
    
			
    2. 
						
  2. NSNotificationCenter详解
			
    本文转载至 http://blog.csdn.net/chengyingzhilian/article/details/7874408 作用:NSNotificationCenter是专门供程序中不同 ...
    
			
    3. 
						
  3. [原创]webpack动态设置css路径
			
    在程序入口的最上方添加 __webpack_public_path__ = path; //your path //your app start here
    
			
    4. 
						
  4. 九度OJ 1196：成绩排序 （排序）
			
    时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:4339 解决:1476 题目描述: 用一维数组存储学号和成绩,然后,按成绩排序输出. 输入: 输入第一行包括一个整数N(1<=N< ...
    
			
    5. 
						
  5. python  login form
			
    import time from selenium import webdriver browser = webdriver.Chrome() wait_time = 1 USER = 'xl.fen ...
    
			
    6. 
						
  6. 我的Android进阶之旅------> Android为TextView组件中显示的文本添加背景色
			
    通过上一篇文章 我的Android进阶之旅------> Android在TextView中显示图片方法 (地址:http://blog.csdn.net/ouyang_peng/article ...
    
			
    7. 
						
  7. activiti 基础搭建
			
    首先在eclipse内添加activiti的插件,https://blog.csdn.net/qq_22701869/article/details/79537971 1.创建一个maven的java ...
    
			
    8. 
						
  8. BAPI LIST
			
    [转自 http://blog.csdn.net/minsenwu/article/details/8432081] 库存管理BAPI 库存: 1. BAPI_MATERIAL_AVAILABILIT ...
    
			
    9. 
						
  9. SMARTFORMS 字段格式化设置
			
    [转自http://lz357502668.blog.163.com/blog/static/16496743201273153434564/] 在SMARTFORM 输出的时候有时候会遇到数字类型无 ...
    
			
    10. 
						
  10. ADO.NET概述
			
    xml这类文件它是.net变成环境中优先使用的数据访问借口. ADO.NET传输的数据都是XML格式的 ADO.NET是一组用于和数据源惊醒交互的面向对象类库 数据源:通常是各种数据库,但文本.exc ...
    
			
    11.