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

给定一个二叉树，节点的值仅限于从0-9，每一个从根节点达到叶子节点的路径代表一个数字。

一个例子，如果根节点到叶子节点的路径是 1->2->3，那么代表这个数字是123。

寻找所有路径代表的数字的和。

例如：

    1

   / \

  2   3

从根节点到叶子节点的路径是 1->2 代表的数字是 12.
从根节点到叶子节点的路径是 1->3 代表的数字是  13.

返回和为 sum = 12 + 13 = 25.

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

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

For example,

    1

   / \

  2   3

The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
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) {}
 * };
 */
int sumNumbers(TreeNode *root, int last)
{
    /*没进入一层都要把last乘以10*/
    if(root->left == NULL && root->right == NULL)
    {
        return last * 10 + root->val;
    }
    else if(root->left == NULL)
    {
        return sumNumbers(root->right, last * 10 + root->val);
    }
    else if(root->right == NULL)
    {
        return sumNumbers(root->left, last * 10 + root->val);
    }
    else
    {
        return sumNumbers(root->left, last * 10 + root->val) + sumNumbers(root->right, last * 10 + root->val);
    }
}
int sumNumbers(TreeNode *root)
{
    /*空树*/
    if(root == NULL)
    {
        return 0;
    }
    else
    {
        return sumNumbers(root, 0);
    }
}

// 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);

//81 + 8627 + 8624 + 869 = 18201
    cout << sumNumbers(pNodeA1) << endl;

DestroyTree(pNodeA1);
    return 0;
}

 
结果输出：
18201
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:
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#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);
    }
}

 




 






 


 


 


 


 


 


 


 


 


 


 


 


 


 


 
												


											
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