[LeetCode] Path Sum II 解题报告

最新推荐文章于 2021-11-23 19:25:16 发布

原创最新推荐文章于 2021-11-23 19:25:16 发布 · 166 阅读

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本文介绍了一种解决二叉树路径和问题的方法，通过递归搜索找到所有从根节点到叶子节点且路径和等于给定数值的路径。

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]
]

» Solve this problem

[解题报告]
二叉树递归。

[Code]

1:    vector<vector<int> > pathSum(TreeNode *root, int sum) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      vector<vector<int> > collect;  
5:      vector<int> solution;  
6:      if(root!=NULL)  
7:        GetPath(root, sum, 0, solution, collect);  
8:      return collect;  
9:    }  
10:    void GetPath(TreeNode* node, int sum, int cal, vector<int>& solution, vector<vector<int> >& collect)  
11:    {   
12:      solution.push_back(node->val);  
13:      cal += node->val;  
14:      if(cal == sum && node->left == NULL && node->right == NULL)  
15:      {  
16:        collect.push_back(solution);        
17:      }  
18:      else  
19:      {      
20:        if(node->left != NULL)  
21:        {        
22:          GetPath(node->left, sum, cal, solution, collect);        
23:        }  
24:        if(node->right != NULL)  
25:        {        
26:          GetPath(node->right, sum, cal, solution, collect);  
27:        }  
28:      }  
29:      solution.pop_back();  
30:      cal -= node->val;  
31:      return;  
32:    }