98. Validate Binary Search Tree
Given the root of a binary tree, determine if it is a valid binary search tree (BST).
A valid 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:
Input: root = [2,1,3]
Output: true
Example 2:
Input: root = [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.
Constraints:
- The number of nodes in the tree is in the range [1, 104].
- -231 <= Node.val <= 231 - 1
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean isValidBST(TreeNode root) {
if(root==null){
return true;
}
return dfs(root,Long.MIN_VALUE,Long.MAX_VALUE);
}
boolean dfs(TreeNode root,long min,long max){
// 当前节点为空直接返回true
if(root == null){
return true;
}
// 当前节点比min或者left节点小，或者当前节点比right节点大，返回false
if(root.val<=min||root.val>=max){
return false;
}
// 递归执行(left，min, 根结点) &&(right，根结点, max)
return dfs(root.left,min,(int)root.val)&&
dfs(root.right,(int)root.val,max);
}
}