【二叉树的递归】04找出二叉树中路径和等于给定值的所有路径【Path Sum II】

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

给定一个二叉树和一个和，判断这个树中是否有一个从根到叶子的路径，使其这个路径上面的所有节点值的和为这个给定的值。

并且返回所有等于给定值的路径。

例如：

给定下面的二叉树，并且和为22。

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

返回true，因为这里面存在一个根到叶子的路径 5->4->11->2，使其他们的和为22。

[
   [5,4,11,2],
   [5,8,4,5]
]

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

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

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

test.cpp:

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#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) {}
* };
*/
void findPathSum(TreeNode *root, vector<int> &path, vector<vector<int> > &allpath, int sum)
{
    /*没访问一个节点就放进去*/
    path.push_back(root->val);
    if(root->left == NULL && root->right == NULL)
    {
        vector<int>::iterator it = path.begin();
        int tmpsum = 0;
        for(; it != path.end(); ++it)
        {
            tmpsum += *it;
        }
        if(tmpsum == sum)
        {
            allpath.push_back(path);
        }
        /*到叶子节点了，而且已经计算过了，就把这个节点退出，并且递归结束，这种节点是叶子节点的情况*/
        path.pop_back();
        return ;
    }

    if(root->left != NULL)
    {
        findPathSum(root->left, path, allpath, sum);
    }
    if(root->right != NULL)
    {
        findPathSum(root->right, path, allpath, sum);
    }
    /*如果这个节点的左右节点都访问过了，把这个节点退出，并且返回上层递归，这种节点是左右不同时空的情况的节点*/
    path.pop_back();
}

vector<vector<int> > pathSum(TreeNode *root, int sum)
{

    vector<vector<int> > allpath;
    if(root == NULL)
    {
        return allpath;
    }
    vector<int> path;
    findPathSum(root, path, allpath, sum);
    return allpath;
}

// 树中结点含有分叉，
//                  8
//              /       \
//             6         12
//           /   \
//          9     2
//               / \
//              4   7
int main()
{
    TreeNode *pNodeA1 = CreateBinaryTreeNode(8);
    TreeNode *pNodeA2 = CreateBinaryTreeNode(6);
    TreeNode *pNodeA3 = CreateBinaryTreeNode(12);
    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 = pathSum(pNodeA1, 20);

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

    DestroyTree(pNodeA1);
    return 0;
}

结果输出：

8 6 2 4

8 12

BinaryTree.h:

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#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:

#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);
    }
}

posted @ 2014-04-08 16:59 z陵阅读(417) 评论(0) 编辑收藏举报

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【二叉树的递归】04找出二叉树中路径和等于给定值的所有路径【Path Sum II】

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