669. Trim a Binary Search Tree 修剪二叉搜索树


作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/


@TOC

题目地址:https://leetcode.com/problems/trim-a-binary-search-tree/description/

题目描述

Given a binary search tree and the lowest and highest boundaries as L and R, trim the tree so that all its elements lies in [L, R] (R >= L). You might need to change the root of the tree, so the result should return the new root of the trimmed binary search tree.

Example 1:

Input: 
    1
   / \
  0   2

  L = 1
  R = 2

Output: 
    1
      \
       2

Example 2:

Input: 
    3
   / \
  0   4
   \
    2
   /
  1

  L = 1
  R = 3

Output: 
      3
     / 
   2   
  /
 1

题目大意

给定[L,R]区间,进行BST裁剪。只要数值不在该区间的节点,全部都删除。返回的结果应该仍然是个BST.

解题方法

递归

想法很简单了,如果root的值比R大,说明root以及其所有右节点都不在这个区间内,向左边搜索。如果root的值比L小,,说明root以及其所有左节点都不在这个区间内,向右边搜索。如果root正好在这个区间内,那么分别对它的左右子树进行裁剪就好了。

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

class Solution:
    def trimBST(self, root, L, R):
        """
        :type root: TreeNode
        :type L: int
        :type R: int
        :rtype: TreeNode
        """
        if not root:
            return None
        if root.val > R:
            return self.trimBST(root.left, L, R)
        elif root.val < L:
            return self.trimBST(root.right, L, R)
        else:
            root.left = self.trimBST(root.left, L, R)
            root.right = self.trimBST(root.right, L, R)
            return root

日期

2018 年 11 月 8 日 —— 项目进展缓慢