題目：

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
 

解釋： 先中序遍歷取出來,還原成一棵樹的時候把加過的值替換上去(還原的時候從右邊開始還原，中序遍歷的時候從左邊開始取)。

python程式碼：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def convertBST(self, root):
 

        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        middle_list=[]
        def middle(root):
            if root==None:
                return None
            middle(root.left)
            middle_list.append(root.val)
            middle(root.right)
        middle(
 
root)
        self.s=0
        def middle2tree(root):
            if root==None:
                return None
            middle2tree(root.right)
            self.s+=middle_list.pop()
            root.val=self.s
            middle2tree(root.left)
        middle2tree(root)
        return root

其實用一遍遍歷就可以了，但是要先遍歷右子樹再遍歷左子樹。 python程式碼：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def convertBST(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        self.cur=0
        def dfs(root):
            if not root:
                return
            dfs(root.right)
            self.cur+=root.val
            root.val=self.cur
            dfs(root.left)
        dfs(root)
        return root

c++程式碼：

/**
 * 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:
    int cur=0;
    TreeNode* convertBST(TreeNode* root) {
        dfs(root);
        return root;
    }
    void dfs(TreeNode* root)
    {
        if (!root)
            return ;
        dfs(root->right);
        cur+=root->val;
        root->val=cur;
        dfs(root->left);
    }
};

總結： 正如之前的一道題目，可以在遍歷樹的時候處理一些問題，不需要先把list寫好再處理了。