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Suppose we have a binary tree, and we also have two numbers a and b, we have to find the value of the lowest node that has a and b as descendants. We have to keep in mind that a node can be a descendant of itself.

So, if the input is like

a = 6, b = 2, then the output will be 4

To solve this, we will follow these steps −

Define a method solve() this will take root and a, b

if root is null, then

return -1

if value of root is either a or b, then

return value of root

left := solve(left of root, a, b)

right := solve(right of root, a, b)

if left or right is not -1, then

return value of root

return left if left is not same as -1 otherwise right

from the main method call solve(root)

Let us see the following implementation to get better understanding −

class TreeNode: def __init__(self, val, left=None, right=None): self.val = val self.left = left self.right = right class Solution: def solve(self, root, a, b): if not root: return -1 if root.val in (a, b): return root.val left = self.solve(root.left, a, b) right = self.solve(root.right, a, b) if -1 not in (left, right): return root.val return left if left != -1 else right ob = Solution() root = TreeNode(3) root.left = TreeNode(10) root.right = TreeNode(4) root.right.left = TreeNode(8) root.right.right = TreeNode(2) root.right.left.left = TreeNode(6) print(ob.solve(root, 6, 2))

root = TreeNode(3) root.left = TreeNode(10) root.right = TreeNode(4) root.right.left = TreeNode(8) root.right.right = TreeNode(2) root.right.left.left = TreeNode(6) 6, 2

4

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