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:
Input:
2
/ \
1 3
Output:
true
Example 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.

Analysis

3
/ \
1 5
/ \
2 6
考虑到如上图情形,BST其实要求的是left subtree的所有node的值都比root node的值小,right subtree所有的node的值都比root node的值大,所以上图中的BST并不valid,因为 2 > 3
对于每个node而言,其取值有一个区间,由parent node和grand parent node共同确定,比如上图中的情形,则node 5的left subtree取值范围为开区间(3, 5)
可以定义一个递归的辅助函数:
boolean isValidSubtree (TreeNode root, Integer min, Integer max)
传入min,max来限定子树的取值区间

Solution

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isValidBST(TreeNode root) {
if (root == null) return true;
return isValidSubtree(root, null, null);
}
boolean isValidSubtree (TreeNode root, Integer min, Integer max) {
if (root == null) return true;
if ((min != null && root.val <= min) || (max != null && root.val >= max)) return false;
return isValidSubtree (root.left, min, root.val) && isValidSubtree(root.right, root.val, max);
}
}
Time Complexity -- O(n)
Space Complexity - O(n) (recursive call stack)