You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  \
    5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/path-sum-iii
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我一直以题代练，以练代学，没咋学习基础知识，遇到不会的题都是直接研究别人的代码（这不是抄别人答案的理由~）

今天做的题各种内存错误，说明我自己平常写代码，很少处理边缘数据，数组大小溢出等等，需要多练习吧~

 sums.resize(maxDepth(root) + 1, 0);申请内存忘了加一，报错也看不懂~一大片红 我枯了

/**
 * 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:
    vector<int>sums;
    int count=0;
    void travel(TreeNode * root ,int level ,int sum){
        if(root==NULL){return ;}
        sums[level] = sums[level-1]+root->val;
        for(int i=0;i<level;i++){
            if(sums[level]-sums[i]==sum)count++;
        }
        travel(root->left,level+1,sum);
        travel(root->right,level+1,sum);
    }
    int pathSum(TreeNode* root, int sum) {
        if(root == NULL){return 0;}
        sums.resize(maxDepth(root)+1,0);
        travel(root,1,sum);
        return count;
    }
    int maxDepth(TreeNode * root){
        if(root==NULL)return 0;
        return max(maxDepth(root->left),maxDepth(root->right))+1;
    }
   
};