描述

Given the root of 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.

As a reminder, a binary search tree is a tree that satisfies these constraints:

• The left subtree of a node contains only nodes with keys less than the node’s key.
• The right subtree of a node contains only nodes with keys greater than the node’s key.
• Both the left and right subtrees must also be binary search trees.

Note: This question is the same as 538: https://leetcode.com/problems/convert-bst-to-greater-tree/

Example 1:

Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]

Example 2:

Input: root = [0,null,1]
Output: [1,null,1]

Example 3:

Input: root = [1,0,2]
Output: [3,3,2]

Example 4:

Input: root = [3,2,4,1]
Output: [7,9,4,10]

Note:

The number of nodes in the tree is in the range [1, 100].
0 <= Node.val <= 100
All the values in the tree are unique.
root is guaranteed to be a valid binary search tree.

解析

根据题意，其实就是将搜索二叉树的每个节点变为大于等于其自身的累加和，所以就是先右子树——后根结点——再左子树的顺序，开始遍历所有节点，并将遍历到当前节点的值加入 sum 中，并将 sum 赋值给当前节点即可。

解答

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

运行结果

Runtime: 24 ms, faster than 32.29% of Python online submissions for Binary Search Tree to Greater Sum Tree.
Memory Usage: 13.4 MB, less than 64.57% of Python online submissions for Binary Search Tree to Greater Sum Tree.

原题链接：https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/

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