538. Convert BST to Greater Tree

题目描述：

Given a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST.

Example:

Input: The root of a Binary Search Tree like this:
              5
            /   \
           2     13

Output: The root of a Greater Tree like this:
             18
            /   \
          20     13

解题思路：

起初的思路时间复杂度不好，借鉴别人的思路，DFS是个更好的方法。

先对右儿子进行处理，对值进行累加，再对父节点进行处理，最后对左儿子进行处理。

代码：

起初的代码:

 1 /**
 2  * Definition for a binary tree node.
 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     TreeNode* convertBST(TreeNode* root) {
13         if (root == NULL)
14             return NULL;
15         root->right = convertBST(root->right);
16         if (root->right != NULL)
17             root->val = root->val + findLeft(root->right);
18         if (root->left != NULL) {
19             TreeNode* tmp = findR(root->left);
20             tmp->val = tmp->val + root->val;
21         }
22         root->left = convertBST(root->left);
23         /*if (root->left != NULL)
24             root->left->val = root->left->val + root->val;*/
25         return root;
26     }
27     int findLeft(TreeNode* root) {
28         while (root->left != NULL) {
29             root = root->left;
30         }
31         return root->val;
32     }
33     TreeNode* findR(TreeNode* root) {
34         while (root->right != NULL) {
35             root = root->right;
36         }
37         return root;
38     }
39 };
View Code

DFS：

 1 /**
 2  * Definition for a binary tree node.
 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     TreeNode* convertBST(TreeNode* root) {
13         int sum = 0;
14         DFS(root, sum);
15         return root;
16     }
17     void DFS(TreeNode* root, int& sum) {
18         if (!root) 
19             return;
20         DFS(root->right, sum);
21         sum += root->val;
22         root->val = sum;
23         DFS(root->left, sum);
24     }
25 };
View Code

 

posted @ 2018-08-31 10:32  gszzsg  阅读(145)  评论(0)  编辑  收藏  举报