98. 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:

    2
   / \
  1   3

Binary tree [2,1,3], return true.

 

Example 2:

    1
   / \
  2   3

Binary tree [1,2,3], return false.

Fancy code，参考了discussion

思路：一个valid binary search tree, root肯定要处于Integer.MAX_VALUE和Integer.MIN_VALUE之间，不然无法形成bst.然后再检查左右子树。

/**
 * 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) {
        return dfs(root,Long.MIN_VALUE, Long.MAX_VALUE);
    }
    public boolean dfs(TreeNode root,long min,long max)
    {
        if(root==null)
        {
            return true;
        }
        if(root.val>=max||root.val<=min)
        {
            return false;
        }
       return dfs(root.left,min,root.val)&&dfs(root.right,root.val,max);
        

    }
}

method 2。FROM DISCUSSION

用全局变量记录之前的root，之前的root的left node就是现在的root。

public class Solution {
    TreeNode pre = null;
    public boolean isValidBST(TreeNode root) {
        return inOrder(root);
    }
    
    public boolean inOrder(TreeNode root){
        if(root==null) return true;
        if(!inOrder(root.left)) return false;
        if(pre==null){
            pre= new TreeNode(root.val);
        }else{
            if(root.val<=pre.val)
                return false;
            pre.val=root.val;
        }
        return inOrder(root.right);
    }
}