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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.
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool execute(TreeNode* root,int sum){ sum-=root->val; if(root->left==NULL&&root->right==NULL){ if(sum==0) return true; } if(root->left&&execute(root->left,sum)){ return true; } if(root->right&&execute(root->right,sum)){ return true; } return false; } bool hasPathSum(TreeNode* root, int sum) { if(root==NULL) return false; return execute(root,sum); } };
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原文地址：http://www.cnblogs.com/michaelzhao10/p/4887640.html
