首页 > 代码库 > [LeetCode] 98. Validate Binary Search Tree Java
[LeetCode] 98. Validate Binary Search Tree Java
题目:
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:
2 / 1 3
Binary tree [2,1,3]
, return true.
Example 2:
1 / 2 3
Binary tree [1,2,3]
, return false.
题意及分析:给出一课书,要求判断该树是不是二叉搜索树。二叉搜索树按照中序遍历得到的是一个升序序列,那么这道题只需要对树进行中序遍历即可,使用stack保存中间结果。这里需要注意的是理论最小值的获取,这里先获取最小值,然后当遍历到这个点时不需要做判断。
代码:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public boolean isValidBST(TreeNode root) { if(root==null||(root.left==null&&root.right==null)) return true; TreeNode minNode = root; while(minNode.left!=null){ minNode=minNode.left; } long nowMax=minNode.val; //找到理论的最小值点 Stack<TreeNode> stack = new Stack<>(); TreeNode node = root; stack.add(node); while(node.left!=null||!stack.isEmpty()){ if(node.left!=null){ //一直找到最左子节点 node = node.left; stack.add(node); }else{ //输出该点,对栈当前点的右节点做相同操作 TreeNode now = stack.pop(); if(now!=minNode){ if(now.val>nowMax){ //如果当前大于遍历的上一个那么当前最大值编程当前点最大值 nowMax = now.val; }else{ //按照中序遍历输出,结果当前输出比上一个数小,那么直接返回false return false; } } if(now.right!=null) { node = now.right; stack.add(node); //将当前点加入stack中 } } } return true; } }
[LeetCode] 98. Validate Binary Search Tree Java
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。