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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / 
 2   3
    /
   4
    
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

 

BST如果进行inorder 的话应该是sorted。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int per = 0;
    boolean sig;
    public boolean isValidBST(TreeNode root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        per = Integer.MIN_VALUE;
        sig = true;
        isValid(root);
        return sig;
    }
    public void isValid(TreeNode root){
        if(root == null) return;
        isValid(root.left);
        if(root.val > per) per = root.val;
        else{
            sig = false;
            return;
        }
        isValid(root.right);
    }
}

还是可以使用递归来做：

public class Solution {
    public boolean isValidBST(TreeNode root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        return root == null || check(root.left, Integer.MIN_VALUE, root.val) && check(root.right, root.val, Integer.MAX_VALUE);
    }
    public boolean check(TreeNode root, int min, int max){
        if(root == null) return true;
        if(!(min < root.val && root.val < max)) return false;
        return check(root.left, min, root.val) && check(root.right, root.val, max);
    }
}

这个方法很好！