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Program to create linked list to binary search tree in Python
Suppose we have a sorted linked list node of size n, we have to create a binary search tree by Taking the value of the k = floor of (n / 2) the smallest setting it as the root. Then recursively constructing the left subtree using the linked list left of the kth node. And recursively constructing the right subtree using the linked list right of the kth node.
So, if the input is like [2,4,5,7,10,15], then the output will be
To solve this, we will follow these steps−
Define a method solve(), this will take node
if node is null, then
return null
if next of node is null, then
return a new tree node with value of node
slow := node, fast := node
prev := None
while fast and next of fast are not null, do
prev := slow
slow := next of slow
fast := next of next of fast
next of prev := None
root := a new tree node with value of slow
left of root := solve(node)
right of root := solve(next of slow)
return root
Let us see the following implementation to get better understanding −
Example
class ListNode: def __init__(self, data, next = None): self.val = data self.next = next class TreeNode: def __init__(self, data, left = None, right = None): self.data = data self.left = left self.right = right def make_list(elements): head = ListNode(elements[0]) for element in elements[1:]: ptr = head while ptr.next: ptr = ptr.next ptr.next = ListNode(element) return head def print_tree(root): if root is not None: print_tree(root.left) print(root.data, end = ', ') print_tree(root.right) class Solution: def solve(self, node): if not node: return None if not node.next: return TreeNode(node.val) slow = fast = node prev = None while fast and fast.next: prev = slow slow = slow.next fast = fast.next.next prev.next = None root = TreeNode(slow.val) root.left = self.solve(node) root.right = self.solve(slow.next) return root ob = Solution() head = make_list([2,4,5,7,10,15]) root = ob.solve(head) print_tree(root)
Input
[2,4,5,7,10,15]
Output
2, 4, 5, 7, 10, 15,
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