# Leetcode 236. Lowest Common Ancestor of a Binary Tree

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# 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`

and`q`

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.**