【leetcode】98. Validate Binary Search Tree

最新推荐文章于 2025-06-18 20:53:42 发布

ghscarecrow

最新推荐文章于 2025-06-18 20:53:42 发布

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本文探讨了如何判断一棵二叉树是否为有效的二叉搜索树。通过递归方式检查每个节点，确保左子树的所有节点小于根节点，右子树的所有节点大于根节点，并且左右子树也必须是二叉搜索树。

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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:

Input:
    2
   / \
  1   3
Output: true

Example 2:

    5
   / \
  1   4
     / \
    3   6
Output: false
Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value
             is 5 but its right child's value is 4.

思考：判断一棵二叉树是否为有效的二叉搜索树，对于该题，我们可以使用自顶向下的递归方式进行判断。再对其左子树，右子树进行判断。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        if(root==null){
            return true;
        }
        return isBST(root,null,null);
    }
    
    public boolean isBST(TreeNode root,Integer lower_limit,Integer upper_limit){
        if((lower_limit != null) && (root.val <= lower_limit)){
            return false;
        }
        if((upper_limit != null) && (root.val >= upper_limit)){
            return false;
        }
        boolean left = root.left != null? isBST(root.left,lower_limit,root.val):true;
        if(left) {
            boolean right = root.right != null? isBST(root.right,root.val,upper_limit):true;
            return right;
        } else {
            return false;
        }
    }
}