# Lowest Common Ancestor

## Method 1: Compare Paths From Root

Following is simple O(n) algorithm to find LCA of n1 and n2.

- Find path from root to n1 and store it in a vector or array.
- Find path from root to n2 and store it in another vector or array.
- Traverse both paths till the values in arrays are same. Return the common element just before the mismatch.

Time Complexity: Time complexity of the above solution is O(n). The tree is traversed twice, and then path arrays are compared.

## Method 2: Using Single Traversal

The method 1 finds LCA in O(n) time, but requires three tree traversals plus extra spaces for path arrays. If we assume that the keys n1 and n2 are present in Binary Tree, we can find LCA using single traversal of Binary Tree and without extra storage for path arrays. The idea is to traverse the tree starting from root. If any of the given keys (n1 and n2) matches with root, then root is LCA (assuming that both keys are present). If root doesn’t match with any of the keys, we recur for left and right subtree. The node which has one key present in its left subtree and the other key present in right subtree is the LCA. If both keys lie in left subtree, then left subtree has LCA also, otherwise LCA lies in right subtree.

```
struct Node *findLCA(struct Node* root, int n1, int n2)
{
// Base case
if (root == NULL) return NULL;
// If either n1 or n2 matches with root's key, report
// the presence by returning root (Note that if a key is
// ancestor of other, then the ancestor key becomes LCA
if (root->key == n1 || root->key == n2)
return root;
// Look for keys in left and right subtrees
Node *left_lca = findLCA(root->left, n1, n2);
Node *right_lca = findLCA(root->right, n1, n2);
// If both of the above calls return Non-NULL, then one key
// is present in once subtree and other is present in other,
// So this node is the LCA
if (left_lca && right_lca) return root;
// Otherwise check if left subtree or right subtree is LCA
return (left_lca != NULL)? left_lca: right_lca;
}
```

Time Complexity: Time complexity of the above solution is O(n) as the method does a simple tree traversal in bottom up fashion.

Note that the above method assumes that keys are present in Binary Tree. If one key is present and other is absent, then it returns the present key as LCA (Ideally should have returned NULL).

We can extend this method to handle all cases by passing two boolean variables v1 and v2. v1 is set as true when n1 is present in tree and v2 is set as true if n2 is present in tree.

## TODO

LCA in binary search tree