Leetcode::Pathsum & Pathsum II

Pathsum

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.

分析：二叉树root-to-leaf的查找，深搜搜到结果返回即可

 /**
  * Definition for binary tree
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
     bool hasPathSum(TreeNode *root, int sum) {
         if(root==NULL) return false;
         int sumval = root->val;
         if(sumval==sum && root->left==NULL &&root->right==NULL) return true;
         else{
             return (hasPathSum(root->left,sum-sumval)||hasPathSum(root->right,sum-sumval));
         }
     }
 };

PathsumII

Description:

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]
]
分析：这一题和上一题的区别在于不仅需要判断有没有，还需要记录是什么。在深搜的时候维护一个stack（这里用vector实现），记录已经走过的路径，
返回是栈也弹出相应节点

 /**
  * Definition for binary tree
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
     bool pathrec(vector<vector<int> > &rec, vector<int>& path,TreeNode* r,int sum)
     {
         sum-=r->val;
         //if(sum<0) return false;
         //Tell if it's a leaf node
         if(r->left ==NULL && r->right == NULL)
         {
             if(sum!=) return false;
             else{
                 vector<int> one = path;
                 one.push_back(r->val);
                 rec.push_back(one);
                 return true;
             }
         }
         //Ordinary node
         path.push_back(r->val);
         bool flag = false;
         if(r->left!=NULL)  pathrec(rec,path,r->left,sum);
         if(r->right!=NULL)  pathrec(rec,path,r->right,sum);
         path.erase(path.end()-);
         return flag;

     }
     vector<vector<int> > pathSum(TreeNode *root, int sum) {
         vector<vector<int> > rec;
         if(root==NULL) return rec;
         vector<int> path;
         pathrec(rec,path,root,sum);

         return rec;
     }
 };