> For the complete documentation index, see [llms.txt](https://ondrej-kvasnovsky-2.gitbook.io/algorithms/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://ondrej-kvasnovsky-2.gitbook.io/algorithms/data-structures/binary-tree/lowest-common-ancestor-of-a-binary-tree.md).

# Lowest Common Ancestor of a Binary Tree

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](https://en.wikipedia.org/wiki/Lowest_common_ancestor): “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**).”

Given the following binary search tree: root = \[3,5,1,6,2,0,8,null,null,7,4]

```
        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4
```

**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 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.
```

**Note:**

* All of the nodes' values will be unique.
* p and q are different and both values will exist in the binary tree.

## Solution

We need to find case when a node equals p or q. So we go from leafs up.

```
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public static TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        //base case
        if (root == null || root == p || root == q) {
            return root;
        }
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);

        //result
        if(left == null) {
            return right;
        }
        else if(right == null) {
            return left;
        }
        else { //both left and right are not null, we found our result
            return root;
        }
    }
}
```