Find Lowest Common Ancestor of a Binary Tree 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 of. Also, a particular node can also be a descendant of itself. We have to find the node and return it as an output.

So, if the input is like

and x = 2, y = 4; then the output will be 3.

The node of which the nodes 2 and 4 are descendant of is 3. So, 3 will be returned.

To solve this, we will follow these steps −

• Define a function dfs() . This will take node

  • if node is similar to null, then

    • return

  • if node is present in list [x,y], then

    • left := dfs(left of node)

    • right := dfs(right of node)

    • if left or right is non-zero, then

      • ans := node

      • return node

  • left := dfs(left of node)

  • right := dfs(right of node)

  • if left and right is not null, then

    • ans := node

    • return node

  • return left or right

• ans := dfs(root)

• return ans

Example

 Live Demo

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

Input

make_tree([5, 3, 7, 2, 4, 1, 7, 6, 8, 10]),
search_node(root, 2),
search_node(root, 4)

Output

3