终于到了二叉树。题目如下：

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.

 首先这个树很奇怪。。。。我也不知道他是怎么插进去的值的，不像传统二叉树那样有大小判断插入，而是好像在。。。随机插入。。。当然这个我们不管，题目是要问找一条从根到叶的路径加下来sum能否等于给的sum。

 如果二叉树基本功熟练的话，可以直接看出来这就是一个深度优先的搜索，然后这个是前序遍历加个和就行了。解法如下：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:void PreSum(TreeNode *temp, int sum, int tmp, bool &flag){	if (temp != NULL)	{		if (temp->val + tmp == sum && temp->left == NULL && temp->right == NULL)			flag = true;		else		{			tmp += temp->val;			PreSum(temp->left, sum, tmp, flag);			PreSum(temp->right, sum, tmp, flag);		}		}	}    bool hasPathSum(TreeNode *root, int sum) {        bool flag = false;        PreSum(root, sum, 0, flag);        return flag;    }};

 因为这个递归弹出实在没法跟要求一样，所以我加了个flag作为判断。

[leetcode] 4. Path Sum