# Lowest Common Ancestor of Deepest Leaves in Python

PythonServer Side ProgrammingProgramming

Suppose we have a a rooted binary tree, we have to return the lowest common ancestor of its deepest leaves. We have to keep in mind that −

• The node of a binary tree is a leaf node if and only if it has no children

• The depth of the root of the tree is 0, and when the depth of a node is d, the depth of each of its children is d+1.

• The lowest common ancestor of a set S of nodes is the node A with the largest depth such that every node in S is in the subtree with root A.

If the input is [1,2,3,4,5],

then the output will be [2,4,5]

To solve this, we will follow these steps −

• Define a method called solve(), this will take node, this will work as follows −

• if node is not present, then return a list with [0, None]

• if left and right subtrees are empty of node, then return a list with [1, None]

• d1, l := solve(left of node), d2, r := solve(right of node)

• if d1 > d2 , then return a list with values [d1 + 1, l]

• otherwise when d2 > d1, then return a list with values [d2 + 1, r]

• return a list with values [d1 + 1, node]

• In the main method, we will perform −

• list := solve(root)

• return list[1]

## Example

Let us see the following implementation to get better understanding −

class Solution(object):
"""
:type root: TreeNode
:rtype: TreeNode
"""
return self.solve(root)[1]
return temp
def solve(self,node):
if not node:
return [0,None]
if not node.left and not node.right:
return [1,node]
d1,l = self.solve(node.left)
d2,r = self.solve(node.right)
if d1>d2:
return [d1+1,l]
elif d2>d1:
return [d2+1,r]
return [d1+1,node]

## Input

[1,2,3,4,5]

## Output

[2,4,5]
Published on 17-Mar-2020 06:54:49