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.

Approach #1: C++.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode* root) {
        vector<int> inorderTraverseArray;
        inorder(root, inorderTraverseArray);
        for (int i = 1; i < inorderTraverseArray.size(); ++i) {
            if (inorderTraverseArray[i-1] >= inorderTraverseArray[i]) return false;
        }
        return true;
    }
    
private:
    void inorder(TreeNode* root, vector<int>& inorderTraverseArray) {
        if (root == NULL) return ;
        if (root->left != NULL) inorder(root->left, inorderTraverseArray);
        inorderTraverseArray.push_back(root->val);
        if (root->right != NULL) inorder(root->right, inorderTraverseArray);
    }
};

　　

Approach #2: Java.

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

　　

Approach #3: Python.

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        output = []
        self.solve(root, output)
            
        for i in range(1, len(output)):
            if output[i-1] >= output[i]:
                return False
        return True
    
    def solve(self, root, output):
        if root == None:
            return
        self.solve(root.left, output)
        output.append(root.val)
        self.solve(root.right, output)