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)