112. Path Sum（DFS）

1. Description

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.

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2. Analysis

做法和 “257. Binary Tree Paths（DFS）”其实是一样的。只不过这样做有点low，把所有的结果都找出去，再遍历结果查找是否存在路径上结点总和为sum的路径，复杂度为O(n)，相当于遍历整个树。其实可以在遍历过程就结束查找的，只不过有bug还没解决，待续。

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3. code

比较low的解法：

/** * 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:    static void recursizeBTP(TreeNode* root, int count, vector<int>& res) {        if(root == NULL) return;        count += root->val;        if(root->left == NULL && root->right == NULL) {                res.push_back(count);                return;        }        recursizeBTP(root->left, count, res);        recursizeBTP(root->right, count, res);        return;    }    bool hasPathSum(TreeNode* root, int sum) {        if(root == NULL) return false;         if(root->left == NULL && root->right == NULL) {            if(root->val == sum) return true;            else return false;        }        int count = root->val;        vector<int> res;        TreeNode * node = root;        if(root->left != NULL) {            recursizeBTP(root->left, count, res);        }        if(root->right != NULL) {            recursizeBTP(root->right, count, res);        }        for(int i = 0; i < res.size(); i++) {            if(res[i] == sum)              return true;        }        return false;    }};