LeetCode | 0538. 把二叉搜索树转换为累加树【Python】

旧街凉风 提交于 2021-01-07 22:09:01

Problem

LeetCode

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 1038: https://leetcode.com/problems/binary-search-tree-to-greater-sum-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]

Constraints:

  • The number of nodes in the tree is in the range [0, 104].
  • -104 <= Node.val <= 104
  • All the values in the tree are unique.
  • root is guaranteed to be a valid binary search tree.

问题

力扣

给出二叉 搜索 树的根节点,该树的节点值各不相同,请你将其转换为累加树(Greater Sum Tree),使每个节点 node 的新值等于原树中大于或等于 node.val 的值之和。

提醒一下,二叉搜索树满足下列约束条件:

  • 节点的左子树仅包含键 小于 节点键的节点。
  • 节点的右子树仅包含键 大于 节点键的节点。
  • 左右子树也必须是二叉搜索树。

**注意:**本题和 1038: https://leetcode-cn.com/problems/binary-search-tree-to-greater-sum-tree/ 相同

示例 1:

输入:[4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
输出:[30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]

示例 2:

输入:root = [0,null,1]
输出:[1,null,1]

示例 3:

输入:root = [1,0,2]
输出:[3,3,2]

示例 4:

输入:root = [3,2,4,1]
输出:[7,9,4,10]

提示:

  • 树中的节点数介于 0 和 104 之间。
  • 每个节点的值介于 -104 和 104 之间。
  • 树中的所有值 互不相同 。
  • 给定的树为二叉搜索树。

思路

中序遍历

利用 BST 的中序遍历就是升序的特性,降序遍历 BST 的元素值。

Python3 代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def convertBST(self, root: TreeNode) -> TreeNode:
        def dfs(root):
            nonlocal sumval
            if root:
                dfs(root.right)
                sumval += root.val
                root.val = sumval  # 将BST转化成累加树
                dfs(root.left)
        
        sumval = 0
        dfs(root)
        return root

GitHub 链接

Python

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