[Leetcode][Tree][Path Sum I & II]

很简单的一道题，判断和为sum的路径是否存在。

结果WA了两次，一次是由于没有考虑清楚深搜停止的条件，另外一次是由于没有考虑到Path必须是从root到leaf的。

 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) {
14             return false;
15         } else {
16             if (sum == root->val && root->left == NULL && root->right == NULL) {
17                 return true;
18             }
19             sum -= root->val;
20             return hasPathSum(root->left, sum) || hasPathSum(root->right, sum);
21         }
22     }
23 };

第一次遇到跟题解写的思路一模一样的情况，但是题解还是比我写的更精简，更高效，减少了root==null时候的调用。

 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) {
14             return false;
15         } else {
16             if (root->left == NULL && root->right == NULL) {
17                 return sum == root->val;
18             }
19             sum -= root->val;
20             return hasPathSum(root->left, sum) || hasPathSum(root->right, sum);
21         }
22     }
23 };

 

另外一个题Path Sum 2

就是把结果保存下来罢了，常见做法。

 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         res.clear();
14         vector<int> tmp;
15         dfs(root, sum, tmp);
16         return res;
17     }
18 private:
19     vector<vector<int> > res;
20     void dfs(TreeNode * root, int sum, vector<int> tmp) {
21         if (root == NULL) {
22             return;
23         }
24         tmp.push_back(root->val);
25         sum -= root->val;
26         if (root->left == NULL && root->right == NULL && sum == 0) {
27             res.push_back(tmp);
28         }
29         dfs(root->left, sum, tmp);
30         dfs(root->right, sum, tmp);
31     }
32 };

看了一下题解， 有一个非常关键的地方，我调用的vector参数没有用引用&，会造成空间复杂度非常大，应该避免这种做法。

 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         res.clear();
14         vector<int> tmp;
15         dfs(root, sum, tmp);
16         return res;
17     }
18 private:
19     vector<vector<int> > res;
20     void dfs(TreeNode * root, int sum, vector<int>& tmp) {
21         if (root == NULL) {
22             return;
23         }
24         tmp.push_back(root->val);
25         sum -= root->val;
26         if (root->left == NULL && root->right == NULL && sum == 0) {
27             res.push_back(tmp);
28         }
29         dfs(root->left, sum, tmp);
30         dfs(root->right, sum, tmp);
31         tmp.pop_back();
32     }
33 };

 

 

posted @ 2014-07-05 16:39  poemqiong  阅读(174)  评论(0)  编辑  收藏  举报