LeetCode 98.Validate Binary Search Tree

xiaoxiao2021-02-28  6

LeetCode 98.Validate Binary Search Tree

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.

方法1:利用本身性质:左<根<右,初始时带入系统的最大值和最小值,在递归过程中换成它们自己的值,;

public boolean isValidBST(TreeNode root) { if(root==null) return true; return valid(root, Long.MIN_VALUE, Long.MAX_VALUE); } public boolean valid(TreeNode root,long low,long high){ if(root==null) return true; if(root.val<=low||root.val>=high) return false; return valid(root.left, low, root.val)&&valid(root.right, root.val, high); }

方法2:将二叉树进行中序遍历,若遍历结果是递增的有序序列,则是合法的二叉搜索树。

public boolean isValidBST(TreeNode root) { List<Integer> res=new ArrayList<>(); InOrderTree(res, root); for(int i=0;i<res.size()-1;i++){ if(res.get(i)>=res.get(i+1)) return false; } return true; } public void InOrderTree(List<Integer> list,TreeNode root){ if(root!=null){ InOrderTree(list, root.left); list.add(root.val); InOrderTree(list, root.right); } }
转载请注明原文地址: https://www.6miu.com/read-1750170.html

最新回复(0)