mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-28 22:51:54 +08:00
99 lines
2.2 KiB
Markdown
Executable File
99 lines
2.2 KiB
Markdown
Executable File
# [98. Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)
|
||
|
||
|
||
## 题目
|
||
|
||
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 **greater than** the node's key.
|
||
- Both the left and right subtrees must also be binary search trees.
|
||
|
||
**xample 1:**
|
||
|
||
2
|
||
/ \
|
||
1 3
|
||
|
||
Input: [2,1,3]
|
||
Output: true
|
||
|
||
**Example 2**:
|
||
|
||
5
|
||
/ \
|
||
1 4
|
||
/ \
|
||
3 6
|
||
|
||
Input: [5,1,4,null,null,3,6]
|
||
Output: false
|
||
Explanation: The root node's value is 5 but its right child's value is 4.
|
||
|
||
## 题目大意
|
||
|
||
给定一个二叉树，判断其是否是一个有效的二叉搜索树。假设一个二叉搜索树具有如下特征：
|
||
|
||
- 节点的左子树只包含小于当前节点的数。
|
||
- 节点的右子树只包含大于当前节点的数。
|
||
- 所有左子树和右子树自身必须也是二叉搜索树。
|
||
|
||
|
||
## 解题思路
|
||
|
||
- 判断一个树是否是 BST，按照定义递归判断即可
|
||
|
||
|
||
## 代码
|
||
|
||
```go
|
||
|
||
package leetcode
|
||
|
||
import "math"
|
||
|
||
/**
|
||
* Definition for a binary tree node.
|
||
* type TreeNode struct {
|
||
* Val int
|
||
* Left *TreeNode
|
||
* Right *TreeNode
|
||
* }
|
||
*/
|
||
|
||
// 解法一，直接按照定义比较大小，比 root 节点小的都在左边，比 root 节点大的都在右边
|
||
func isValidBST(root *TreeNode) bool {
|
||
return isValidbst(root, math.Inf(-1), math.Inf(1))
|
||
}
|
||
func isValidbst(root *TreeNode, min, max float64) bool {
|
||
if root == nil {
|
||
return true
|
||
}
|
||
v := float64(root.Val)
|
||
return v < max && v > min && isValidbst(root.Left, min, v) && isValidbst(root.Right, v, max)
|
||
}
|
||
|
||
// 解法二，把 BST 按照左中右的顺序输出到数组中，如果是 BST，则数组中的数字是从小到大有序的，如果出现逆序就不是 BST
|
||
func isValidBST1(root *TreeNode) bool {
|
||
arr := []int{}
|
||
inOrder(root, &arr)
|
||
for i := 1; i < len(arr); i++ {
|
||
if arr[i-1] >= arr[i] {
|
||
return false
|
||
}
|
||
}
|
||
return true
|
||
}
|
||
|
||
func inOrder(root *TreeNode, arr *[]int) {
|
||
if root == nil {
|
||
return
|
||
}
|
||
inOrder(root.Left, arr)
|
||
*arr = append(*arr, root.Val)
|
||
inOrder(root.Right, arr)
|
||
}
|
||
|
||
``` |