LCA Of Binary Tree
You have been given a Binary Tree of distinct integers and two nodes ‘X’ and ‘Y’. You are supposed to return the LCA (Lowest Common Ancestor) of ‘X’ and ‘Y’.

The LCA of ‘X’ and ‘Y’ in the binary tree is the shared ancestor of ‘X’ and ‘Y’ that is located farthest from the root.

Note :

You may assume that given ‘X’ and ‘Y’ definitely exist in the given binary tree.

For example :

For the given binary tree

LCA of ‘X’ and ‘Y’ is highlighted in yellow colour.

Input Format :

The first line contains an integer 'T' which denotes the number of test cases to be run. Then the test cases follows The first line of each test case contains elements of the tree in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place. The second line of each test case contains two integers ‘X’ and ‘Y’ denoting the two nodes of the binary tree. For example, the input for the tree depicted in the below image would be :

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Explanation :

Level 1 : The root node of the tree is 1 Level 2 : Left child of 1 = 2 Right child of 1 = 3 Level 3 : Left child of 2 = 4 Right child of 2 = null (-1) Left child of 3 = 5 Right child of 3 = 6 Level 4 : Left child of 4 = null (-1) Right child of 4 = 7 Left child of 5 = null (-1) Right child of 5 = null (-1) Left child of 6 = null (-1) Right child of 6 = null (-1) Level 5 : Left child of 7 = null (-1) Right child of 7 = null (-1) The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on. The input ends when all nodes at the last level are null(-1).

Note :

The above format was just to provide clarity on how the input is formed for a given tree. The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as: 1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Output Format :

For each test case, print an integer denoting the LCA of the given binary tree nodes. For each test case, print the output in a separate line.

Note :

You do not need to print anything, it has already been taken care of. Just implement the given function.

Constraints :

1

Question

LCA Of Binary Tree

You have been given a Binary Tree of distinct integers and two nodes ‘X’ and ‘Y’. You are supposed to return the LCA (Lowest Common Ancestor) of ‘X’ and ‘Y’.

The LCA of ‘X’ and ‘Y’ in the binary tree is the shared ancestor of ‘X’ and ‘Y’ that is located farthest from the root.

Note :

You may assume that given ‘X’ and ‘Y’ definitely exist in the given binary tree.

For example :

For the given binary tree

Example

LCA of ‘X’ and ‘Y’ is highlighted in yellow colour.

Input Format :

The first line contains an integer 'T' which denotes the number of test cases to be run. Then the test cases follows

The first line of each test case contains elements of the tree in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place.

The second line of each test case contains two integers ‘X’ and ‘Y’ denoting the two nodes of the binary tree.

For example, the input for the tree depicted in the below image would be :

Example

1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1

Explanation :

Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 2
Right child of 1 = 3

Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6

Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)

Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null(-1).

Note :

The above format was just to provide clarity on how the input is formed for a given tree. 
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Output Format :

For each test case, print an integer denoting the LCA of the given binary tree nodes.

For each test case, print the output in a separate line.

Note :

You do not need to print anything, it has already been taken care of. Just implement the given function.

Constraints :

1 <= 'T' <= 100
0 <= 'N' <= 3000
0 <= 'DATA' <= 10^4

Where 'DATA' is a node value of the binary tree.

Time Limit: 1 sec

Accepted Answer

Naive Approach

The basic idea of this approach is to list down all the ancestors of the given nodes. And then, we will choose the common ancestor from the two lists which is farthest from the root of the tree.

Consider the binary tree as shown in the above figure, where we are trying to find the LCA of X and Y. Let us try to find the path from the root node to X and Y respectively and store the nodes present in the path in two separate lists pathToX and pathToY. Observe that the first few nodes(here nodes in yellow colour) in the lists will be the same which are common ancestors of node X and Y. Now, we need a common ancestor which is farthest from the root node. It is quite clear from the above image that the last common node in the lists will be the LCA.

Consider the following steps to describe the approach explained above:

Initialise two empty lists pathToX and pathToY to store the nodes in the path from the root node to ‘X’ and ‘Y’ respectively.
- Perform a pre-order traversal to find the nodes in the path from the root node to X and store it in the pathToX such that the first element in the list is the root node.
- Similarly, perform a pre-order traversal to find the nodes in the path from the root node to ‘Y’ and store it in the pathToY such that the first element in the list is the root node.
Now, start iterating from the front of the lists while the elements are equal.
The last equal element will be the LCA of X and Y.

Space Complexity: O(n)Explanation:

O(N), Where N is the number of nodes in the given binary tree.

Since we are storing the paths in a list and in the worst case (Skewed Trees), all nodes of the given tree can be present in the list when ‘X’ will be the root node and ‘Y’ will be the leaf node of the binary tree. So the overall space complexity will be O(N).

Time Complexity: O(n)Explanation:

O(N), Where N is the number of nodes in the given binary tree.

Since we are doing two pre-order traversals and in the worst case (Skewed Trees), all nodes of the given tree can be present in the path when ‘X’ and ‘Y’ are very close to the leaf node of the binary tree. And each one can take O(N) time. So the overall time complexity will be O(N).

Java (SE 1.8)

/*

    Time Complexity - O(N)

    Space Complexity - O(N)



    Where 'N' is the number of nodes in the Binary Tree.

*/



import java.util.ArrayList;



public class Solution 

{

	//    Finds the path from root to 'X' and return true if it exists.

	private static boolean getPath(TreeNode root, ArrayList path, int x) 

	{

		if (root == null) 

		{

			//    Tree doesn't have any node.

			return false;

		}



		//    Store the root node in the path

		path.add(root.data);



		//    Check if we reached to the destination node i.e. 'X'

		if (root.data == x) 

		{

			//    'X' is found.

			return true;

		}



		//    Recur to find 'X' in the left subtree.

		if (getPath(root.left, path, x)) 

		{

			//    'X' exists in the left subtree.

			return true;

		}



		//    Recur to find 'X' in the right subtree.

		if (getPath(root.right, path, x)) 

		{

			//    'X' exists in the right subtree.

			return true;

		}



		//    Remove the root node because 'X' doesn't exists in the subtree rooted at "root".

		path.remove(path.size() - 1);



		return false;

	}





	public static int lowestCommonAncestor(TreeNode root, int x, int y) 

	{

		//    To store the path from the root node to 'X' and 'Y'

		ArrayList pathToX = new ArrayList();

		ArrayList pathToY = new ArrayList();



		//    Get the path from the root node to node 'X'

		getPath(root, pathToX, x);



		//    Get the path from the root node to node 'Y'

		getPath(root, pathToY, y);



		int index;



		//    Iterate while elements in the lists are equal.

		for (index = 0; index < Math.min(pathToX.size(), pathToY.size()); index++) 

		{

			if (pathToX.get(index) != pathToY.get(index)) 

			{

				//    Elements are not same.

				break;

			}

		}



		//    Return the LCA.

		return pathToX.get(index - 1);

	}

}

Python (3.5)

'''
    Time Complexity : O(N)
    Space Complexity : O(N)

    Where 'N' is the number of nodes in the given binary tree.\
'''


def getPath(root, path, x):
    # Base Case
    if root == None:
        return False
    path.append(root.data)

    # Check if we reached to the destination node i.e. 'X'
    if (root.data == x):
        return True

    # Recur to find 'X' in the left and the right subtree.
    if getPath(root.left, path, x) or getPath(root.right, path, x):
        return True

    # Remove the root node because 'X' doesn't exists in the subtree rooted at "root".
    del path[-1]

    return False


def lowestCommonAncestor(root, x, y):
    # To store the path from the root node to 'X' and 'Y'
    pathToX, pathToY = [], []
    # Get the path from the root node to node 'X'
    getPath(root, pathToX, x)
    # Get the path from the root node to node 'Y'
    getPath(root, pathToY, y)

    # Iterate while elements in the lists are equal.
    for index in range(min(len(pathToX), len(pathToY))):
        if pathToX[index] != pathToY[index]:
            # If elements are not same, return the LCA
            return pathToX[index - 1]

    # Return the LCA.
    return pathToX[min(len(pathToX), len(pathToY)) - 1]

C++ (g++ 5.4)

/*
    Time Complexity : O(N)
    Space Complexity : O(N)

    Where 'N' is the number of nodes in the given binary tree.
*/

//    Finds the path from root to 'X' and return true if it exists.
bool getPath(TreeNode *root, vector &path, int x)
{

	if (root == NULL)
	{
		//    Tree doesn't have any node.
		return false;
	}

	//    Store the root node in the path
	path.push_back(root->data);

	//    Check if we reached to the destination node i.e. 'X'
	if (root->data == x)
	{
		//    'X' is found.
		return true;
	}

	//    Recur to find 'X' in the left subtree.
	if (getPath(root->left, path, x))
	{
		//    'X' exists in the left subtree.
		return true;
	}

	//    Recur to find 'X' in the right subtree.
	if (getPath(root->right, path, x))
	{
		//    'X' exists in the right subtree.
		return true;
	}

	//    Remove the root node because 'X' doesn't exists in the subtree rooted at "root".
	path.pop_back();

	return false;
}

int lowestCommonAncestor(TreeNode *root, int x, int y)
{
	//    To store the path from the root node to 'X' and 'Y'
	vector pathToX, pathToY;

	//    Get the path from the root node to node 'X'
	getPath(root, pathToX, x);

	//    Get the path from the root node to node 'Y'
	getPath(root, pathToY, y);

	int index;

	//    Iterate while elements in the lists are equal.
	for (index = 0; index < min(pathToX.size(), pathToY.size()); index++)
	{
		if (pathToX[index] != pathToY[index])
		{
			//    Elements are not same.
			break;
		}
	}

	//    Return the LCA.
	return pathToX[index - 1];
}

Accepted Answer

Space Optimised Approach

The basic idea of this approach is to find the LCA in a single traversal without using any extra space.

Let us start moving from the root node. Now consider the following situations:

If either X or Y is the root node, then the LCA of X and Y will be the root node itself because the root node is the topmost node in the binary tree.
If X and Y exist in the left subtree, then the LCA will be necessarily present in the left subtree because the farthest common ancestor from the root node will be present in the left subtree.
Similarly, if X and Y exist in the right subtree, then the LCA will be necessarily present in the right subtree.
If both X and Y are present in the different subtrees, then the LCA will be the root node.

We can easily generalize the points mentioned above for any node. Let

findLCA(TreeNode* root, int X, int Y)

be a function that returns the LCA of X and Y in the given tree or return -1 if it does not exist. Now consider the steps as follows:

If the root is NULL, then return -1 because LCA will not exist in an empty tree.
If either the root is equal to X or Y return root. In this case, the root node itself will be LCA because the root is the topmost node in the binary tree.
Let us find the LCA in the left subtree and store it in a variable leftLCA, i.e. leftLCA = findLCA(left-child of the root, X, Y)
Similarly, find the LCA in the right subtree and store it in a variable rightLCA i.e. rightLCA = findLCA(right-child of the root, X, Y)
If leftLCA and rightLCA both are not equal to -1, then the root must be the LCA. Because both ‘X’ and ‘Y’ are present in two different subtrees.
Otherwise, the LCA must be present in either of the two subtrees.
- If leftLCA is not equal to -1, return leftLCA.
- Otherwise, return rightLCA.

Space Complexity: O(n)Explanation:

O(N), Where N is the number of nodes in the given binary tree.

Since we are doing a recursive tree traversal and in the worst case (Skewed Trees), all nodes of the given tree can be stored in the call stack. So the overall space complexity will be O(N).

Time Complexity: O(n)Explanation:

O(N), Where N is the number of nodes in the given binary tree.

Since we are doing a pre-order traversal which takes O(N) time. So the overall time complexity will be O(N).

Python (3.5)

'''

    Time Complexity : O(N)

    Space Complexity : O(N)



    Where 'N' is the number of nodes in the given binary tree.

'''





# Returns the LCA of X and Y or return -1 if it does not exist.

def findLCA(root, x, y):



    # Base Case

    if root == None:

        return -1

    # The root node itself is a LCA

    elif root.data == x or root.data == y:

        return root.data

    else:

        # Recur to find the LCA in the left subtree

        leftLCA = findLCA(root.left, x, y)

        # Recur to find the LCA in the right subtree

        rightLCA = findLCA(root.right, x, y)

        if leftLCA != -1 and rightLCA != -1:

            # The root must be the LCA since both X and Y are present in two different subtrees.

            return root.data

        elif leftLCA != -1:

            # LCA exists in the left subtree.

            return leftLCA

        else:

            # LCA exists in the right subtree.

            return rightLCA





def lowestCommonAncestor(root, x, y):

    return findLCA(root, x, y)

Java (SE 1.8)

/*

    Time Complexity - O(N)

    Space Complexity - O(N)



    Where 'N' is the number of nodes in the Binary Tree.

*/



public class Solution 

{

    //    Returns the LCA of 'X' and 'Y'or return -1 if it does not exist.

    private static int findLCA(TreeNode root, int x, int y) 

    {

        if (root == null) 

        {

            //    LCA can't exist in an empty tree.

            return -1;

        } 

        else if (root.data == x || root.data == y) 

        {

            //   The root node itself will be the LCA.

            return root.data;

        } 

        else 

        {

            //    Recur to find the LCA in the left subtree.

            int leftLCA = findLCA(root.left, x, y);



            //    Recur to find the LCA in the right subtree.

            int rightLCA = findLCA(root.right, x, y);



            if (leftLCA != -1 && rightLCA != -1) 

            {

                //    The root must be the LCA since both 'X' and 'Y' are present in two different subtrees.

                return root.data;

            } 

            else if (leftLCA != -1) 

            {

                //    LCA exists in the left subtree.

                return leftLCA;

            } 

            else 

            {

                //    LCA exists in the right subtree.

                return rightLCA;

            }

        }

    }



    public static int lowestCommonAncestor(TreeNode root, int x, int y) 

    {

        int lca = findLCA(root, x, y);

        return lca;

    }

}

C++ (g++ 5.4)

/*
    Time Complexity : O(N)
    Space Complexity : O(N)
	
    Where 'N' is the number of nodes in the given binary tree.
*/

//    Returns the LCA of ‘X’ and ‘Y’ or return -1 if it does not exist.
int findLCA(TreeNode *root, int x, int y)
{

	if (root == NULL)
	{
		//    LCA can't exist in an empty tree.
		return -1;
	}
	else if (root->data == x || root->data == y)
	{
		//   The root node itself will be the LCA.
		return root->data;
	}
	else
	{
		//    Recur to find the LCA in the left subtree.
		int leftLCA = findLCA(root->left, x, y);

		//    Recur to find the LCA in the right subtree.
		int rightLCA = findLCA(root->right, x, y);

		if (leftLCA != -1 && rightLCA != -1)
		{
			//    The root must be the LCA since both ‘X’ and ‘Y’ are present in two different subtrees.
			return root->data;
		}
		else if (leftLCA != -1)
		{
			//    LCA exists in the left subtree.
			return leftLCA;
		}
		else
		{
			//    LCA exists in the right subtree.
			return rightLCA;
		}
	}
}

int lowestCommonAncestor(TreeNode *root, int x, int y)
{
	return findLCA(root, x, y);
}

You have been given a Binary Tree of distinct integers and two nodes ‘X’ and ‘Y’. You are supposed to return the LCA (Lowest Common Ancestor) of ‘X’ and ‘Y’.

The LCA of ‘X’ and ‘Y’ in the binary tree is the shared ancestor of ‘X’ and ‘Y’ that is located farthest from the root.

Note :

For example :

Input Format :

Explanation :

Note :

Output Format :

Note :

Constraints :

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