Leetcode-98: Validate Binary Search Tree
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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 3Binary tree
[2,1,3]
, return true.Example 2:
1 / \ 2 3Binary tree
[1,2,3]
, return false.验证一颗二叉树是否是搜索二叉树。(不允许重复键的二叉树)
标签:深度优先搜索。
思路:对于二叉树的深度优先搜索实际上就是先序遍历(preorder traversal)和中序遍历(inorder traversal),两者的入栈和出栈顺序是一样的,区别在于:遍历到节点A时,先序遍历先访问节点A,再遍历A的左子节点;中序遍历则先遍历左子节点,出栈后再访问节点A。
由于对于搜索二叉树来说,有一个特殊的性质,即中序遍历后的结果是升序的(这里需要注意,如果允许重复键的话,搜索二叉树要求左子树的键不大于根节点的键,此时无法直接用中序遍历来验证是否是搜索二叉树,因为无法区分l<= root < r和 l < root <= r)。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */class Solution { private long prev = (long)Integer.MIN_VALUE - 1; public boolean isValidBST(TreeNode root) { return inorder(root); } private boolean inorder(TreeNode h) { if (h == null) return true; if (!inorder(h.left)) return false; if (h.val > prev) prev = h.val; else return false; return inorder(h.right); }}
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