[leetcode]98. Validate Binary Search Tree验证二叉搜索树

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.

Solution1: 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

    5  (min, max)                             5 :in(min,max)?
   /                                 /   / return true               return false
  1   4                           1 update(min, 5), in (min, 5)?    4 update(5, 4), in (5,4)?

code

/**
 * 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) {
                           /*why using null, null? 因为对于root，没有办法确定范围*/      
       return helper(root, null, null);
    }
                                  /* 因为上面定义的是null，这里initialize就应该是Integer而不是int*/   
    private boolean helper(TreeNode root, Integer min, Integer max) {
        // base case 
        if (root == null)return true;
        
        // based on BST attribute
        if ((min != null && root.val <= min) || (max != null && root.val >= max))
            return false;
        
        // dfs
        Boolean left = helper(root.left, min, root.val);
        Boolean right = helper(root.right, root.val, max);
        return  left && right;
    }
}

复杂度

Time: O(n) : visit each node 

Space: O(h):  h is the deepest height of such tree