Search in a Binary Search Tree

Question

Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node's value equals the given value. Return the subtree rooted with that node. If such node doesn't exist, you should return NULL.

For example,

Given the tree:
        4
       / \
      2   7
     / \
    1   3

And the value to search: 2
You should return this subtree:

  2     
 / \   
1   3

In the example above, if we want to search the value 5, since there is no node with value 5, we should return NULL.

Note that an empty tree is represented by NULL, therefore you would see the expected output (serialized tree format) as [], not null.

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

class Solution:
    def searchBST(self, root: TreeNode, val: int) -> TreeNode:


        if(root is None) or (root.val == val):
            return root

        if(root.val>val):
            return self.searchBST(root.right,val)
        elif(root.val<val):
            return self.searchBST(root.left,val)

This is the code that I have written,

Now if I change the second if condition to :

return self.searchBST(root.left, val) if val < root.val \
            else self.searchBST(root.right, val)

The code is working fine, HELP?

Answer 1

You have flipped the values. If root.val > val you need to search root.left.

class Solution:
    def searchBST(self, root: TreeNode, val: int) -> TreeNode:


        if(root is None) or (root.val == val):
            return root

        if(root.val>val):
            return self.searchBST(root.left,val) // FLIPPED
        elif(root.val<val):
            return self.searchBST(root.right,val) // FLIPPED

Answer 2

Your code is not totally correct, if the current node value (root) is greater than the value you are searching for, then go left, otherwise go right.

In your current code, you are doing the opposite.

So, the correct way would be like this:

if(root.val>val):
        return self.searchBST(root.left,val) #changed to root.left
    elif(root.val<val):
        return self.searchBST(root.right,val) #changed to root.right