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.
Input: 2 / 1 3 Output: true
题意:
验证BST
思路:
dfs
根据BST的性质: 左子树<根<右子树
要特别留心,对于input的root,没有办法给定一个确定的范围。所以初始化为null
代码:
1 class Solution { 2 public boolean isValidBST(TreeNode root) { 3 return helper(root, null, null); 4 } 5 6 boolean helper(TreeNode root, Integer min, Integer max) { 7 if (root == null) 8 return true; 9 10 if ((min != null && root.val <= min) || (max != null && root.val >= max)) 11 return false; 12 13 return helper(root.left, min, root.val) && helper(root.right, root.val, max); 14 } 15 }
原文地址:https://www.cnblogs.com/liuliu5151/p/9175890.html
时间: 2024-10-10 13:37:33