Path Sum--路径和(重)

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问题：链接

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.

解答：

DFS或者BFS。同时，当本节点是叶子节点才判断和，否则节点为空时返回false

参考：http://www.cnblogs.com/remlostime/archive/2012/11/13/2767746.html

代码：

class Solution {public:    bool hasPathSum(TreeNode *root, int sum) {return DFS(root, 0, sum);    }bool DFS(TreeNode *root, int sum, int target){if(root == NULL)return false;if(root->left == NULL && root->right == NULL)return (target == sum + root->val);return DFS(root->left, sum+root->val, target) || DFS(root->right, sum+root->val, target);}};

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