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100 lines
3.0 KiB
Markdown
100 lines
3.0 KiB
Markdown
# [669. Trim a Binary Search Tree](https://leetcode.com/problems/trim-a-binary-search-tree/)
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## 题目
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Given the `root` of a binary search tree and the lowest and highest boundaries as `low` and `high`, trim the tree so that all its elements lies in `[low, high]`. Trimming the tree should **not** change the relative structure of the elements that will remain in the tree (i.e., any node's descendant should remain a descendant). It can be proven that there is a **unique answer**.
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Return *the root of the trimmed binary search tree*. Note that the root may change depending on the given bounds.
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**Example 1:**
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```
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Input: root = [1,0,2], low = 1, high = 2
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Output: [1,null,2]
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```
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**Example 2:**
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```
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Input: root = [3,0,4,null,2,null,null,1], low = 1, high = 3
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Output: [3,2,null,1]
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```
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**Example 3:**
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```
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Input: root = [1], low = 1, high = 2
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Output: [1]
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```
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**Example 4:**
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```
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Input: root = [1,null,2], low = 1, high = 3
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Output: [1,null,2]
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```
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**Example 5:**
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```
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Input: root = [1,null,2], low = 2, high = 4
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Output: [2]
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```
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**Constraints:**
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- The number of nodes in the tree in the range `[1, 10^4]`.
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- `0 <= Node.val <= 10^4`
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- The value of each node in the tree is **unique**.
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- `root` is guaranteed to be a valid binary search tree.
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- `0 <= low <= high <= 10^4`
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## 题目大意
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给你二叉搜索树的根节点 root ，同时给定最小边界low 和最大边界 high。通过修剪二叉搜索树，使得所有节点的值在[low, high]中。修剪树不应该改变保留在树中的元素的相对结构（即，如果没有被移除，原有的父代子代关系都应当保留）。 可以证明，存在唯一的答案。所以结果应当返回修剪好的二叉搜索树的新的根节点。注意，根节点可能会根据给定的边界发生改变。
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## 解题思路
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- 这一题考察二叉搜索树中的递归遍历。递归遍历二叉搜索树每个结点，根据有序性，当前结点如果比 high 大，那么当前结点的右子树全部修剪掉，再递归修剪左子树；当前结点如果比 low 小，那么当前结点的左子树全部修剪掉，再递归修剪右子树。处理完越界的情况，剩下的情况都在区间内，分别递归修剪左子树和右子树即可。
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## 代码
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```go
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package leetcode
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import (
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"github.com/halfrost/LeetCode-Go/structures"
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)
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// TreeNode define
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type TreeNode = structures.TreeNode
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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func trimBST(root *TreeNode, low int, high int) *TreeNode {
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if root == nil {
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return root
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}
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if root.Val > high {
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return trimBST(root.Left, low, high)
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}
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if root.Val < low {
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return trimBST(root.Right, low, high)
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}
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root.Left = trimBST(root.Left, low, high)
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root.Right = trimBST(root.Right, low, high)
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return root
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}
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``` |