LeetCode: Path Sum

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LeetCode: Path Sum

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.

地址：https://oj.leetcode.com/problems/path-sum/

演算法：用遞迴應該很簡單吧。直接看代碼吧：

 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)   return false;14         return subSolution(root,sum);15     }16     bool subSolution(TreeNode *root, int sum){17         if(!root->left && !root->right){18             if(sum == root->val)    return true;19             else    return false;20         }else{21             if(root->left){ 22                 bool left = subSolution(root->left,sum - root->val);23                 if(left)    return true;24             }25             if(root->right){26                 bool right = subSolution(root->right,sum - root->val);27                 if(right)   return true;28             }29         }30         return false;31     }32 };

第二題：

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]]
地址：https://oj.leetcode.com/problems/path-sum-ii/
演算法：跟上一題一樣，只不過必須返回所有的結果。應該也很簡單，直接看代碼：

 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     vector<vector<int> > pathSum(TreeNode *root, int sum) {13         if(!root)   return vector<vector<int> >();14         return subPathSum(root,sum);15     }16     vector<vector<int> > subPathSum(TreeNode *root, int sum){17         if(!root->left && !root->right){18             if(root->val == sum){19                 return vector<vector<int> >(1,vector<int>(1,sum)); 20             }else{21                 return vector<vector<int> >();22             }23         }24         vector<vector<int> > result;25         if(root->left){26             vector<vector<int> > left = subPathSum(root->left, sum - root->val);27             vector<vector<int> >::iterator it = left.begin();28             for(; it != left.end(); ++it){29                 vector<int> temp;30                 temp.push_back(root->val);31                 temp.insert(temp.end(),it->begin(),it->end());32                 result.push_back(temp);33             }34         }35         if(root->right){36             vector<vector<int> > right = subPathSum(root->right,sum - root->val);37             vector<vector<int> >::iterator it = right.begin();38             for(; it != right.end(); ++it){39                 vector<int> temp;40                 temp.push_back(root->val);41                 temp.insert(temp.end(),it->begin(),it->end());42                 result.push_back(temp);43             }44         }45         return result;46     }47 };

 

LeetCode: Path Sum