Validate Binary Search Tree

最新推荐文章于 2024-10-14 16:31:29 发布

原创最新推荐文章于 2024-10-14 16:31:29 发布 · 209 阅读

0 ·

CC 4.0 BY-SA版权

算法专栏收录该内容

46 篇文章

订阅专栏

本文介绍了一种通过递归方式检查二叉树是否为有效的二叉搜索树的方法。该方法利用辅助函数递归地检查每个节点及其子树，确保左子树的所有节点值小于根节点值，右子树的所有节点值大于根节点值，并且左右子树也必须是有效的二叉搜索树。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.
A single node tree is a BST

分析：此题需要注意的是一个平衡二叉树的条件为左子树的val中最大的值需要小于root.val，右子树的val最小值需要大于root.val，一旦有任意子树不满足上述条件，则该树不是平衡二叉树。

class ResultType {
    boolean isValid;
    int max;
    int min;
    ResultType(boolean a, int b, int c) {
        isValid = a;
        max = b;
        min = c;
    }
}
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: True if the binary tree is BST, or false
     */
    public boolean isValidBST(TreeNode root) {
        ResultType rt = helper(root);
        return rt.isValid;
    }
    private ResultType helper(TreeNode root) {
        if (root == null) {
            return new ResultType(true, Integer.MIN_VALUE, Integer.MAX_VALUE);
        }
        ResultType left = helper(root.left);
        ResultType right = helper(root.right);
        if (!left.isValid || !right.isValid) {
            return new ResultType(false, 0, 0);
        }
        if ((root.left != null && left.max >= root.val) || (root.right != null && right.min <= root.val)) {
            return new ResultType(false, 0, 0);
        }
        return new ResultType(true,
                              Math.max(root.val, right.max),
                              Math.min(root.val, left.min));
    }
}