Leetcode 236. Lowest Common Ancestor of a Binary Tree
Problem Statement
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p
and q
as the lowest node in T
that has both p
and q
as descendants (where we allow a node to be a descendant of itself).”
Test Cases
Example 1:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.
Example 2:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
Example 3:
Input: root = [1,2], p = 1, q = 2
Output: 1
Constraints
- The number of nodes in the tree is in the range
[2, 105]
. -10^9 <= Node.val <= 10^9
- All
Node.val
are unique. p != q
p
andq
will exist in the tree.
Approach
The lowest common ancestor is defined between two nodes p
and q
as the lowest node in T
that has both p
and q
as descendants (where we allow a node to be a descendant of itself.
So we mainly need to check for some cases.
- Whether our node is already null, if yes then return null.
- Whether our node is equal to p or q. The root will surely be the LCA for p and q. Return node in this case.
- Take the left subtree and right subtree in a recursive way.
left = fun(node.left, p, q) and right = fun(node.right, p, q). - If both left and right are null, we return null.
- If both left and right are not null, we return the node as the root note will be the LCA here.
- Otherwise, we check whether left != null, then return left or right != null, then return right.
Time and Space Complexity
We are traversing the whole tree in the worst case. So if the total number of nodes = n, then time complexity will be O(n) where n = number of nodes.
Space complexity will be O(H) where H = height of the tree.