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     13Output: 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 };

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 };