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:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* convertBST(TreeNode* root) {
        if (root == NULL)
            return NULL;
        root->right = convertBST(root->right);
        if (root->right != NULL)
            root->val = root->val + findLeft(root->right);
        if (root->left != NULL) {
            TreeNode* tmp = findR(root->left);
            tmp->val = tmp->val + root->val;
        }
        root->left = convertBST(root->left);
        /*if (root->left != NULL)
            root->left->val = root->left->val + root->val;*/
        return root;
    }
    int findLeft(TreeNode* root) {
        while (root->left != NULL) {
            root = root->left;
        }
        return root->val;
    }
    TreeNode* findR(TreeNode* root) {
        while (root->right != NULL) {
            root = root->right;
        }
        return root;
    }
};

DFS：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* convertBST(TreeNode* root) {
        int sum = 0;
        DFS(root, sum);
        return root;
    }
    void DFS(TreeNode* root, int& sum) {
        if (!root) 
            return;
        DFS(root->right, sum);
        sum += root->val;
        root->val = sum;
        DFS(root->left, sum);
    }
};