98. 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.
Example 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 #
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)
}