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https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-05 08:27:30 +08:00
Add solution 0665、0669、1423、1463、1579
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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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package leetcode
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import (
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"fmt"
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"testing"
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"github.com/halfrost/LeetCode-Go/structures"
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)
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type question669 struct {
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para669
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ans669
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}
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// para 是参数
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// one 代表第一个参数
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type para669 struct {
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one []int
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low int
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high int
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}
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// ans 是答案
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// one 代表第一个答案
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type ans669 struct {
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one []int
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}
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func Test_Problem669(t *testing.T) {
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qs := []question669{
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{
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para669{[]int{1, 0, 2}, 1, 2},
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ans669{[]int{1, structures.NULL, 2}},
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},
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{
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para669{[]int{3, 0, 4, structures.NULL, 2, structures.NULL, structures.NULL, 1}, 1, 3},
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ans669{[]int{3, 2, structures.NULL, 1}},
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},
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{
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para669{[]int{1}, 1, 2},
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ans669{[]int{1}},
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},
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{
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para669{[]int{1, structures.NULL, 2}, 1, 3},
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ans669{[]int{1, structures.NULL, 2}},
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},
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{
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para669{[]int{1, structures.NULL, 2}, 2, 4},
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ans669{[]int{2}},
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},
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}
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fmt.Printf("------------------------Leetcode Problem 669------------------------\n")
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for _, q := range qs {
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_, p := q.ans669, q.para669
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fmt.Printf("【input】:%v ", p)
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root := structures.Ints2TreeNode(p.one)
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fmt.Printf("【output】:%v \n", structures.Tree2ints(trimBST(root, p.low, p.high)))
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}
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fmt.Printf("\n\n\n")
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}
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100
leetcode/0669.Trim-a-Binary-Search-Tree/README.md
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100
leetcode/0669.Trim-a-Binary-Search-Tree/README.md
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# [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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```
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