Program to find out the node in the right in a binary tree using Python

Suppose, we are provided a binary tree. We are also given a pointer to a node (named ‘u’) and we have to find the node situated just right of the provided node. The node situated to the given node's right must stay at the same level and the given node can either be a leaf node or an internal node.

So, if the input is like

and u = 6, then the output will be 8.

The node situated at the right of node 6 is node 8, so the value 8 is returned to us.

To solve this, we will follow these steps −

  • if root is empty, then

    • return null

  • dq := a new deque

  • insert root at the end of dq

  • while dq is not empty, do

    • dq_size := size of dq

    • temp := a new list

    • index := -1

    • for each value in range 0 to dq_size, do

      • node := delete last element from dq

      • if left of node is not empty, then

        • add left of node to the end of dq

      • if right of node is not empty, then

        • add right of node to the end of dq

      • insert node at the end of temp

      • if node is same as u, then

        • index := size of temp - 1

    • if index is same as size of temp - 1, then

      • return null

    • if index > -1, then

      • return temp[index + 1]

  • return null

Let us see the following implementation to get better understanding −


from queue import deque
class TreeNode:
   def __init__(self, val=0, left=None, right=None):
      self.val = val
      self.left = left
      self.right = right
def insert(temp,data):
   que = []
   while (len(que)):
      temp = que[0]
      if (not temp.left):
         if data is not None:
            temp.left = TreeNode(data)
            temp.left = TreeNode(0)
      if (not temp.right):
         if data is not None:
            temp.right = TreeNode(data)
            temp.right = TreeNode(0)
def make_tree(elements):
   Tree = TreeNode(elements[0])
   for element in elements[1:]:
      insert(Tree, element)
   return Tree
def search_node(root, element):
   if (root == None):
      return None
   if (root.val == element):
      return root
   res1 = search_node(root.left, element)
   if res1:
      return res1
   res2 = search_node(root.right, element)
   return res2
def solve(root, u):
   if not root:
      return None
   dq = deque()
   while dq:
      dq_size = len(dq)
      temp = []
      index = -1
      for _ in range(dq_size):
         node = dq.pop()
         if node.left:
         if node.right:
         if node == u:
            index = len(temp) - 1
      if index == len(temp) - 1:
         return None
      if index > -1:
         return temp[index + 1]
   return None
root = make_tree([5, 3, 7, 2, 4, 6, 8])
u = search_node(root,6)
ret = solve(root, u)


root = make_tree([5, 3, 7, 2, 4, 6, 8])
u = search_node(root,6)