# 236 Lowest Common Ancestor of a Binary Tree – Medium

### Problem:

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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

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

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

### Thoughts:

This is similar to the earlier problem, find Lowest Common Ancestor of a Binary Search Tree. The different is that now it is not a Binary Search Tree, there is no easy way to tell if a node is in the left subtree or in the right subtree. Now we have to iterate the nodes of the tree to determine if a subtree contains a node. The trick here is how to iterate effectively the tree.

We iterate from root to bottom, if current dealing node is either p or q, then that’s the lowest common ancestor. Otherwise, we continue to try with curr.left and curr.right.

### Solutions:

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null) {
return null;
}
if (root == p || root == q) {
return root;
}
TreeNode left = lowestCommonAncestor(root.left, p, q);
TreeNode right = lowestCommonAncestor(root.right, p, q);
if (left != null && right != null) {
return root;
}
if (left != null) {
return left;
}
else {
return right;
}
}
}
``````