Program to find out the lowest common ancestor of a binary tree using parent pointers using Python

Suppose, we are given a binary tree and also two specific nodes x and y. We have to find out the lowest common ancestor of the two nodes from the binary tree. The lowest common ancestor in a binary tree is the lowest node of which both the nodes x and y are descendants. A particular node can also be a descendant of itself. We have to find the node and return it as an output.

The node structure of the tree is like below −

   data: <integer>
   left: <pointer of TreeNode>
   right: <pointer of TreeNode>
   parent: <pointer of TreeNode>

We have to utilize the parent pointer while finding the solution.

So, if the input is like

and x = 3, y = 7; then the output will be 5.

3 and 7 are both descendants of 5, so the answer will be 5.

To solve this, we will follow these steps −

  • path_p_r := a new list

  • while x is not null, do

    • insert x at the end of path_p_r

    • x := parent of x

  • while y is not null, do

    • if y is present in path_p_r, then

      • return y

    • y := parent of y

Let us see the following implementation to get better understanding −


 Live Demo

class TreeNode:
   def __init__(self, data, left = None, right = None, parent = None): = data
      self.left = left
      self.right = right
      self.parent = parent
def insert(temp,data):
   que = []
   while (len(que)):
      temp = que[0]
      if (not temp.left):
         if data is not None:
            temp.left = TreeNode(data, parent=temp)
            temp.left = TreeNode(0,parent=temp)
   if (not temp.right):
      if data is not None:
         temp.right = TreeNode(data,parent=temp)
         temp.right = TreeNode(0,parent=temp)
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 ( == element):
      return root
      res1 = search_node(root.left, element)
   if res1:
      return res1
   res2 = search_node(root.right, element)
   return res2
def solve(x, y):
   path_p_r = []
   while x:
      x = x.parent
   while y:
      if y in path_p_r:
         return y
      y = y.parent
root = make_tree([5, 3, 7, 2, 4, 1, 7, 6, 8, 10])
print(solve(search_node(root, 3), search_node(root, 7)).data)


[5, 3, 7, 2, 4, 1, 7, 6, 8, 10], 3, 7



Updated on: 29-May-2021


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