What is wrong with this iterative solution in validating Binary Search Tree?

Question

The goal is, given the root of a binary tree, determine if it is a valid binary search tree (BST).

A valid 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.

The problem setting is the same as here

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        
        level=[root]
        
        while level:
            for k in level:
                
                if k.left:
                    if k.left >= k.val:
                        return False
                    
                if k.right:
                    if k.right <= k.val:
                        return False
                    
            level=[leaf for n in level for leaf in (n.left,n.right) if leaf]
                    
        return True

The above solution could not work for the case [2,1,3] which I believe the bug is located at this part after some debugging processes

if k.left:
   if k.left >= k.val:
      return False

What is wrong with it? Thanks in advance for any help.

How are you making your tree. It seems like this: `k.left >= k.val` is an error because you are comparing `k.left`, which I assume should be another `TreeNode` to `k.val` which is a number. It *really* helps people answer your question if you give enough code to reproduce your error without filling in the blanks and making assumptions. — Mark, Jul 23 '22 at 15:13
What does "does not work" mean? If this is Python 3, you should get a `TypeError` for trying to compare a `TreeNode` and an (assumed) `int`, rather than silently producing an unexpected Boolean value. — chepner, Jul 23 '22 at 15:34

constantstranger · Accepted Answer · 2022-07-23T15:32:17.093

There are a couple of issues with your code:

In the lines if k.left >= k.val: and if k.right <= k.val: you are comparing nodes to values, which I assume is not your intention.
Assuming the above is corrected, the general logic only compares the values of children with those of their immediate parents, which is not enough to satisfy the part of the problem specification stating, "The left subtree of a node contains only nodes with keys less than the node's key", or the symmetrical part of the spec regarding the right subtree. To satisfy this, you need to track the ancestral bounds with which a given node's val attribute must comply.

Here's an example of a modification to your code that would work as an iterative solution:

# Definition for a binary tree node.
class TreeNode(object):
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right
class Solution(object):
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        
        level=[(root,None,None)]
        
        while level:
            level2 = []
            for k,floor,ceil in level:
                
                if k.left:
                    if k.left.val >= k.val or floor is not None and k.left.val <= floor:
                        return False
                    level2.append((k.left,floor,k.val))
                    
                if k.right:
                    if k.right.val <= k.val or ceil is not None and k.right.val >= ceil:
                        return False
                    level2.append((k.right,k.val,ceil))

                level = level2
                    
        return True

t = TreeNode(2, TreeNode(1), TreeNode(3))
x = Solution()
print(x.isValidBST(t))

Ouput:

True

Explanation:

instead of storing each non-null node at a given level, we store a tuple for each such node including the node itself as well as floor and ceil values (set to None for root) that will allow us to carry forward (or downward as we descend the levels of the tree) the bounds to which a given subtree must conform.
When we traverse leftward via a non-null left attribute, we overwrite ceil with the immediate parent's val attribute, and for rightward traversal via a non-null right attribute we overwrite floor.

What is wrong with this iterative solution in validating Binary Search Tree?

1 Answers1