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

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool hasPathSum(TreeNode *root, int sum) {
13         if(root==NULL) return false;
14         int sumval = root->val;
15         if(sumval==sum && root->left==NULL &&root->right==NULL) return true;
16         else{
17             return (hasPathSum(root->left,sum-sumval)||hasPathSum(root->right,sum-sumval));
18         }
19     }
20 };

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

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool pathrec(vector<vector<int> > &rec, vector<int>& path,TreeNode* r,int sum)
13     {
14         sum-=r->val;
15         //if(sum<0) return false;
16         //Tell if it's a leaf node
17         if(r->left ==NULL && r->right == NULL)
18         {
19             if(sum!=0) return false;
20             else{
21                 vector<int> one = path;
22                 one.push_back(r->val);
23                 rec.push_back(one);
24                 return true;
25             }
26         }
27         //Ordinary node
28         path.push_back(r->val);
29         bool flag = false;
30         if(r->left!=NULL)  pathrec(rec,path,r->left,sum);
31         if(r->right!=NULL)  pathrec(rec,path,r->right,sum);
32         path.erase(path.end()-1);
33         return flag;
34         
35     }
36     vector<vector<int> > pathSum(TreeNode *root, int sum) {
37         vector<vector<int> > rec;
38         if(root==NULL) return rec;
39         vector<int> path;
40         pathrec(rec,path,root,sum);
41         
42         return rec;
43     }
44 };

相关阅读:
前端启动摄像头的API
落谷训练---
树的遍历 (和) 玩转二叉树的总结博客
 L2-010 排座位（并查集）
最长回文（manacher模板）
L2-006 树的遍历
 面试题5：从尾到头打印链表
 面试题4：替换空格
 面试题3：二维数组中的查找
 poj 1511(spfa)
原文地址：https://www.cnblogs.com/soyscut/p/3774526.html