[Leetcode] Path Sum II路径和

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 andsum = 22,

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

return

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

题意：给定一数，在树中找出所有路径和等于该数的情况。
方法一：
使用vector向量实现stack的功能，以方便输出指定路径。思想和代码和Path sum大致相同。

/**
 * 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>> res;
        vector<TreeNode *> vec;
        TreeNode *pre=NULL;
        TreeNode *cur=root;
        int temVal=0;

        while(cur|| !vec.empty())
        {
            while(cur)
            {
                vec.push_back(cur);
                temVal+=cur->val;
                cur=cur->left;
            }
            cur=vec.back();
            if(cur->left==NULL&&cur->right==NULL&&temVal==sum)
            {       //和Path sum最大的区别
                vector<int> temp;
                for(int i=0;i<vec.size();++i)
                    temp.push_back(vec[i]->val);
                res.push_back(temp);
            }
            if(cur->right&&cur->right !=pre)
                cur=cur->right;
            else
            {
                vec.pop_back();
                temVal-=cur->val;
                pre=cur;
                cur=NULL;
            }

        } 
        return res;    
    }
};

方法二：递归法

/**
 * 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>> res;
        vector<int> path;
        findPaths(root,sum,res,path);
        return res;
    }
    void findPaths(TreeNode *root,int sum,vector<vector<int>> &res,vector<int> &path)
    {
        if(root==NULL)  return;
        path.push_back(root->val);
        if(root->left==NULL&&root->right==NULL&&sum==root->val)
            res.push_back(path);
        
        findPaths(root->left,sum-root->val,res,path);
        findPaths(root->right,sum-root->val,res,path);
        path.pop_back();
    }
};