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.
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool hasPathSum(TreeNode *root, int sum) { 13 14 if(root == NULL) 15 return false; 16 17 int tempSum = 0; 18 int flag = 0; 19 pathSum(root, tempSum, sum, flag); 20 if(flag == 1) 21 return true; 22 else 23 return false; 24 } 25 26 void pathSum(TreeNode *root, int tempSum, int sum, int &flag) 27 { 28 if(root != NULL) 29 { 30 if(root->left == NULL && root->right == NULL) 31 { 32 tempSum += root->val; 33 if(tempSum == sum) 34 flag = 1; 35 36 return; 37 } 38 39 else 40 { 41 tempSum += root->val; 42 pathSum(root->left, tempSum, sum, flag); 43 pathSum(root->right, tempSum, sum, flag); 44 } 45 } 46 } 47 };
原文：http://www.cnblogs.com/YQCblog/p/3970191.html