【二叉树的递归】03判断二叉树中有没有和为给定值的路径【Path Sum】

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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。

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。

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Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

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

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

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

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test.cpp：

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#include <iostream> #include <cstdio> #include <stack> #include <vector> #include <queue> #include "BinaryTree.h" using namespace std; bool findPathSum(TreeNode *root, vector<int> &path, int sum) {     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;         }         path.pop_back();         if(tmpsum == sum)         {             return true;         }         else         {             return false;         }     }     bool flag = false;     if(root->left)     {         flag = findPathSum(root->left, path, sum);     }     if(!flag && root->right)     {         flag = findPathSum(root->right, path, sum);     }     path.pop_back();     return flag; } bool hasPathSum(TreeNode *root, int sum) {     if(root == NULL)     {         return false;     }     vector<int> path;     return findPathSum(root, path, sum); } // 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); bool ans = hasPathSum(pNodeA1, 20); if (ans == true)     {         cout << "exist!" << endl;     }     else     {         cout << "not exist!" << endl;     } DestroyTree(pNodeA1);     return 0; }

结果输出：

exist!

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);     } }