【Leetcode】Path Sum II

题目：

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and  sum = 22 ,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

解题思路：用DFS算法进行搜索，搜索到叶子节点并且路径和等于设定值时，将路径存入容器。

代码：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int> > pathSum(TreeNode *root, int sum) {
        vector<vector<int>> Path;
        vector<int> CurrPath;
        build_path(Path, CurrPath, root, sum);
        return Path;
    }
private:
    void build_path(vector<vector<int>> &Path,vector<int> &CurrPath, TreeNode *root, int sum){
        if(!root)return;
        CurrPath.push_back(root->val);
        if(!root->left&&!root->right){
            if(root->val==sum)Path.push_back(CurrPath);
        }else{
            build_path(Path,CurrPath,root->left,sum-root->val);
            build_path(Path,CurrPath,root->right,sum-root->val);
        }
        CurrPath.pop_back();
    }
};