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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
采用递归的方式，C：
/** * Definition for binary tree * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */ bool dfs(struct TreeNode *node, int sum, int cur); bool hasPathSum(struct TreeNode *root, int sum) { return dfs(root, sum, 0); } bool dfs(struct TreeNode *node, int sum, int cur){ if(node == NULL) return false; if(node->left == NULL && node->right == NULL) return cur+node->val == sum; return(dfs(node->left, sum, cur+node->val) || dfs(node->right, sum, cur+node->val)); }
C++：
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool dfs(TreeNode *node, int sum, int curSum) { if (node == NULL) return false; if (node->left == NULL && node->right == NULL) return curSum + node->val == sum; return dfs(node->left, sum, curSum + node->val) || dfs(node->right, sum, curSum + node->val); } bool hasPathSum(TreeNode *root, int sum) { // Start typing your C/C++ solution below // DO NOT write int main() function return dfs(root, sum, 0); } };
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原文地址：http://www.cnblogs.com/zhhc/p/4347187.html
