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

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

给定二叉搜索树,将最低和最高边界作为L和R,修剪树,使其所有元素位于[L,R](R> = L)中。您可能需要更改树的根,所以结果应该返回修剪的二叉搜索树的新根。

  1. class Solution(object):
  2. def trimBST(self, root, L, R):
  3. """
  4. :type root: TreeNode
  5. :type L: int
  6. :type R: int
  7. :rtype: TreeNode
  8. """
  9. if not root:
  10. return root
  11. if root.val < L:
  12. return self.trimBST(root.right, L, R)
  13. if root.val > R:
  14. return self.trimBST(root.left, L, R)
  15. root.left = self.trimBST(root.left, L, R)
  16. root.right = self.trimBST(root.right, L, R)
  17. return root

来自为知笔记(Wiz)

时间: 2024-12-13 11:23:55

669. Trim a Binary Search Tree 修剪二叉搜索树的相关文章

669. Trim a Binary Search Tree修剪二叉搜索树

[抄题]: 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

[CareerCup] 4.5 Validate Binary Search Tree 验证二叉搜索树

4.5 Implement a function to check if a binary tree is a binary search tree. LeetCode上的原理,请参见我之前的博客Validate Binary Search Tree 验证二叉搜索树.

[LeetCode] 255. Verify Preorder Sequence in Binary Search Tree 验证二叉搜索树的先序序列

Given an array of numbers, verify whether it is the correct preorder traversal sequence of a binary search tree. You may assume each number in the sequence is unique. Follow up:Could you do it using only constant space complexity? 给一个数组,验证是否为一个二叉搜索树的

[LeetCode] Validate Binary Search Tree 验证二叉搜索树

Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: 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

[LeetCode] Binary Search Tree Iterator 二叉搜索树迭代器

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. Calling next() will return the next smallest number in the BST. Note: next() and hasNext() should run in average O(1) time and uses

[LeetCode] Recover Binary Search Tree 复原二叉搜索树

Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note:A solution using O(n) space is pretty straight forward. Could you devise a constant space solution? confused what "{1,#,2,3}"

[LeetCode] 99. Recover Binary Search Tree 复原二叉搜索树

Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Example 1: Input: [1,3,null,null,2]   1   /  3     2 Output: [3,1,null,null,2]   3   /  1     2 Example 2: Input: [3,1,4,null,null,2]

173 Binary Search Tree Iterator 二叉搜索树迭代器

实现一个二叉搜索树迭代器.你将使用二叉搜索树的根节点初始化迭代器.调用 next() 将返回二叉搜索树中的下一个最小的数.注意: next() 和hasNext() 操作的时间复杂度是O(1),并使用 O(h) 内存,其中 h 是树的高度. 详见:https://leetcode.com/problems/binary-search-tree-iterator/description/ /** * Definition for binary tree * struct TreeNode { *

Binary Search Tree (二叉搜索树)

一直在看Data Structure and Algorithm Analysis 的原版,英文水平有限看的比较慢.代码功力就更不用说了,所以代码打的还没有看书快……已经在看优先队列了,AVL树还没有打完也是棒棒哒.这会儿就先从二叉树更新开始吧. 二叉树的结构什么的基本都知道,二叉搜索树就是比就简单的二叉树多了一个特性(property)——每个节点的左子叶内的key比节点的key小,而其右子叶的key比节点的key大.这个特性不是唯一的(比如左右子叶相对于其父节点的key值大小顺序可以颠倒),