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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: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.
递归深搜，一类题。解法和Sum Root to Leaf Numbers一样。
1 /** 2 * Definition for a binary tree node. 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 void dfs(TreeNode* root,int sum,int add,bool& has) 13 { 14 add+=root->val; 15 if(root->left==NULL && root->right==NULL) 16 { 17 if(add==sum) 18 { 19 has=true; 20 return; 21 } 22 } 23 if(root->left) dfs(root->left,sum,add,has); 24 if(root->right) dfs(root->right,sum,add,has); 25 26 } 27 bool hasPathSum(TreeNode* root, int sum) { 28 if(!root) return false; 29 int add=0; 30 bool result=false; 31 dfs(root,sum,add,result); 32 return result; 33 } 34 };
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原文地址：http://www.cnblogs.com/Sean-le/p/4814939.html
