LintCode 661 Convert BST to Greater Tree

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.

您在真实的面试中是否遇到过这个题？ 

Yes

样例

Given a binary search Tree ｀{5,2,3}｀:

              5            /   \           2     13

Return the root of new tree

             18            /   \          20     13

思路：

先遍历右子树，再遍历左子树。

"""Definition of TreeNode:class TreeNode:    def __init__(self, val):        self.val = val        self.left, self.right = None, None"""class Solution:    # @param {TreeNode} root the root of binary tree    # @return {TreeNode} the new root    def traverseTree(self, root, prenum):        if root.left == None and root.right == None:            tmp = root.val            root.val += prenum            return tmp        leftnum = 0        rightnum = 0        tmp = root.val        if root.right != None:            rightnum = self.traverseTree(root.right, prenum)        root.val += (rightnum + prenum)        if root.left != None:            leftnum = self.traverseTree(root.left, root.val)        return tmp + leftnum + rightnum            def convertBST(self, root):        # Write your code here        if root == None:            return root        rightnum = 0        if root.right != None:            rightnum = self.traverseTree(root.right, 0)        root.val += rightnum        if root.left != None:            self.traverseTree(root.left, root.val)        return root