Leetcode_98Validate Binary Search Tree

来源：互联网发布：上海编程培训班编辑：程序博客网时间：2024/05/21 16:57

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.

先试图将二叉树中序遍历，将中序遍历的各节点的值存入List中，此时中序遍历后此List中如果该二叉树是BST，那么该List中的顺序其实是递增的，所以只需查看是否有前一个值大于等于后一个值得情况，若存在，就不是BST。

public class Solution {     List<Integer> res = new ArrayList<Integer>();     public boolean isValidBST(TreeNode root) {         if (root == null) return true;         if(root.left == null&&root.right == null) return true;         inorder1(root);         for(int i= 1;i<res.size();i++){             if (res.get(i)<=res.get(i-1)) {                return false;            }         }            return true;        }    private void inorder1(TreeNode treeNode){         if (treeNode.left !=null) {            inorder1(treeNode.left);        }         res.add(treeNode.val);         if (treeNode.right != null) {            inorder1(treeNode.right);        }     }}

0 0