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Problem The Leetcode file system keeps a log each time some user performs a…

August 14, 2020

First try to understand question.

- Its a binary tree, not a binary search tree
- Find first common ancestor
- There is no link to parent nodes

- First, you can search in the tree, whether these two nodes exists in the tree.
- Proceed only when we found both nodes in the tree
- You can find the location of these two nodes on the tree. There can be three possibilities:
- If either element is the actual root, then root is the answer
- Both nodes found on the left OR right side of the root node (same side of root)
- One is on left, other is on right. Roor is the answer.

For Case-2, on finding that both nodes are on the same side. We will repeat our same algorithm going forward in that side only.

Lets see a code for this.

```
public class CommonAncestor {
public static class Node {
public int data;
public Node left;
public Node right;
public Node(int data) {
this.data = data;
this.left = null;
this.right = null;
}
}
/**
* To check whether a node exists in a root element
*/
private boolean covers(Node root, Node toSearch) {
if (root == null || toSearch == null) {
return false;
}
if (root.data == toSearch.data) {
return true;
}
return this.covers(root.left, toSearch) || this.covers(root.right, toSearch);
}
private Node helper1(Node root, Node node1, Node node2) {
if (root == null) {
return null;
}
if (root.data == node1.data || root.data == node2.data) {
return root;
}
boolean firstNodeFoundOnLeft = this.covers(root.left, node1);
boolean firstNodeFoundOnRight = this.covers(root.left, node2);
if (firstNodeFoundOnLeft != firstNodeFoundOnRight) {
//both are on different sides. Found result.
return root;
}
//both are on the same side.
return this.helper1(firstNodeFoundOnLeft ? root.left : root.right,
node1, node2);
}
public Node findCommonAncestor_1(Node root, Node node1, Node node2) {
if (!this.covers(root, node1) || !this.covers(root, node2)) {
//one or both nodes are not in the tree
System.out.println("Not covered");
return null;
}
return this.helper1(root, node1, node2);
}
}
```

- First we check whether our tree covers both the nodes or not. If not, we do not need to go further.
- Then, code starts with
`helper1(root, node1, node2)`

- Some conditions, if our root is null. No need to go further. And, if root element becomes equal to any of our search nodes. This is the result.
- We search for both nodes, on the left side of the root node.
- If both are on different sides. Root is the answer
- Else, we will branch on the side (left or right), where both found.

The function calls recursively, and same code repeats.

Consider the tree to be balanced tree.

- Initially
`covers()`

is called twice. So, its`2 times O(n)`

, which computes to`O(n)`

- Our
`helper1()`

function gets called on branches one time for each node. 2 * n/2 for both nodes. Which comes out to be:`2n/2 + 2n/4 + 2n/8 ~= O(logn)`

Total complexity of this code is `O(n)`

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