Validate Binary Search Tree

最新推荐文章于 2024-10-14 16:31:29 发布

dutsoft

最新推荐文章于 2024-10-14 16:31:29 发布

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本文通过中序遍历二叉搜索树并将结果存入数组来验证其有效性，确保数组为已排序状态。

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    
    public static void copyBST(TreeNode root,List<Integer> list){
        if(root==null) return;
        copyBST(root.left,list);
        list.add(root.val);
        copyBST(root.right,list);
    }
    public boolean isValidBST(TreeNode root) {
        List<Integer> list=new ArrayList<Integer>();
        copyBST(root,list);
        for(int i=1;i<list.size();i++){
            if(list.get(i)<=list.get(i-1)) return false;
        }
        return true;
    }
}

思路：中序遍历二叉搜索数，将每个节点的值存在数组中，则数组是已排序的。