[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