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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:
Input:
2
/
1 3
Output: trueExample 2:
5
/
1 4
/
3 6
Output: false
Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value is 5 but its right child's value is 4.
难度：medium
题目：给定二叉树，判断其是否为二叉搜索树。
思路：递归，并记录当前结点的取值范围。
Runtime: 0 ms, faster than 100.00% of Java online submissions for Validate Binary Search Tree.
Memory Usage: 37.6 MB, less than 100.00% of Java online submissions for Validate Binary Search Tree.
/**
* 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 (null == root) {
return true;
}
return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
public boolean isValidBST(TreeNode root, long minVal, long maxVal) {
if (null == root) {
return true;
}
if (root.val >= maxVal || root.val <= minVal) {
return false;
}
return isValidBST(root.left, minVal, root.val) && isValidBST(root.right, root.val, maxVal);
}
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