Diameter Of Binary Tree
You are given a Binary Tree. You are supposed to return the length of the diameter of the tree.

The diameter of a binary tree is the length of the longest path between any two end nodes in a tree.

Note :

The number of edges between two nodes represents the length of the path between them.

For Example :

For the given binary tree

Nodes in the diameter are highlighted. The length of the diameter, i.e., the path length, is 6.

Input Format :

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

For Example, the input for the tree depicted in the above 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)

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 :

Print an integer that denotes the length of the diameter of the given binary tree. Print the output of each test case in a new line.

Note :

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

Constraints :

1

Question

Diameter Of Binary Tree

You are given a Binary Tree. You are supposed to return the length of the diameter of the tree.

The diameter of a binary tree is the length of the longest path between any two end nodes in a tree.

Note :

The number of edges between two nodes represents the length of the path between them.

For Example :

For the given binary tree

Example

Nodes in the diameter are highlighted. The length of the diameter, i.e., the path length, is 6.

Input Format :

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

tree

For Example, the input for the tree depicted in the above 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)

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 :

Print an integer that denotes the length of the diameter of the given binary tree.

Print the output of each test case in a new line.

Note :

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

Constraints :

1 <= T <= 10
0 <= N <= 10^5
1 <= data <= 10^4

Where ‘N’ is the total number of nodes in the binary tree, and 'data' is the value of the binary tree node.

Time Limit: 1 sec

Accepted Answer

The diameter of a binary tree is the length of the longest path between any two end nodes in the tree.

The diameter of a binary tree can be calculated by finding the maximum of the following three values: 1) the diameter of the left subtree, 2) the diameter of the right subtree, and 3) the longest path that passes through the root node.
To find the diameter of a binary tree, we can use a recursive approach where we calculate the height of each node and update the diameter at each step.
The height of a node is the maximum of the heights of its left and right subtrees plus 1.
The diameter of the binary tree is the maximum of the diameters of its left and right subtrees plus 1.

Accepted Answer

in this i have caluclated the height of left side tree then height of right side tree and added the result and returned the values in recursion. I have created one height function,and then call the function (left_height + right_height + 1) for each node and update the result.

Accepted Answer

Recursion

The basic idea of this approach is to break the problem into subproblems.

Now, there are three possible cases:

The diameter of the tree is present in the left subtree.
The diameter of the tree is present in the right subtree.
The diameter of the tree passes through the root node.

Let us define a recursive function, ‘getDiamter’, which takes the root of the binary tree as input parameter and returns the diameter of the given subtree rooted at the “root” node.

We can easily find the diameter of the left subtree and right subtree by recursion. The main task is to calculate the diameter of the tree corresponding to point 3 mentioned above.

From the above figure, we can observe that if the diameter passes through the root node, then it can be written as the length of the longest path between the leaves which passes through the root. And we can get that using the height of the left and right subtrees.

Now, assume a function, ‘getHeight,’ which returns the height of the subtree rooted at the “root” node.

The longest path length(i.e., Number of edges in the path) between the leaves can be written as:

1 + getHeight(left child of the root node) + getHeight(right child of the root node)

Algorithm

If the ‘root’ node is NULL, return 0. The diameter of an empty tree is 0.
Recur for the left subtree and store the diameter of the left subtree in a variable ‘leftDiameter’, i.e. ‘leftDiameter’ = getDiameter(left child of the root node)
Similarly, recur for the right subtree and store the diameter of the right subtree in a variable ‘rightDiameter’ i.e. ‘rightDiameter’ = getDiameter(right child of the root node)
Now, get the height of the left subtree and right subtree and store it in a variable.
- ‘leftHeight’ = getHeight(left child of the root node)
- ‘rightHeight’ = getHeight(right child of the root node)
The diameter of the given tree will be the maximum of the following terms:
- ‘leftDiameter’
- ‘rightDiameter’
- 1 + ‘leftHeight’ + ‘rightHeight’
Return the maximum of above terms i.e.max(leftDiameter, rightDiameter, 1 + leftHeight + rightHeight).

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^2)Explanation:

O(N^2), Where ‘N’ is the number of nodes in the given binary tree.

We are traversing through every node of the binary time that takes O(N) time, and we are calculating the height for every node, and in the worst case (Skewed Trees), the getHeight() function will take O(N) time. So the overall time complexity will be O(N^2).

C++ (g++ 5.4)

/*

    Time Complexity : O(N^2)

    Space Complexity : O(N)

	

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

*/



int getHeight(TreeNode *root)

{

    if (root == NULL)

    {

        // Height of empty tree is 0.

        return 0;

    }



    // Get the height of left subtree.

    int leftHeight = getHeight(root->left);



    // Get the height of right subtree.

    int rightHeight = getHeight(root->right);



    // Height of the given binary tree will be 1 greater than maximum of "leftHeight" and "rightHeight".

    int height = max(leftHeight, rightHeight) + 1;



    return height;

}



int getDiameter(TreeNode *root)

{

    if (root == NULL)

    {

        // Diameter of an empty tree will be 0.

        return 0;

    }



    // Get the height of left and right subtrees.

    int leftHeight = getHeight(root->left);

    int rightHeight = getHeight(root->right);



    // Recur for left subtree and get the diameter.

    int leftDiameter = getDiameter(root->left);



    // Recur for right subtree and get the diameter.

    int rightDiameter = getDiameter(root->right);



    // Diameter of given binary tree.

    int diameter = max(leftDiameter, max(rightDiameter, leftHeight + rightHeight));



    return diameter;

}



int diameterOfBinaryTree(TreeNode *root)

{

    return getDiameter(root);

}

Python (3.5)

"""
    Time Complexity : O(N^2)
    Space Complexity : O(N)
	
    Where 'N' is the number of nodes in the given binary tree.
"""

def getHeight(root):

    if (root == None):
    
        # Height of empty tree is 0.
        return 0
    

    # Get the height of left subtree.
    leftHeight = getHeight(root.left)

    # Get the height of right subtree.
    rightHeight = getHeight(root.right)

    # Height of the given binary tree will be 1 greater than maximum of "leftHeight" and "rightHeight".
    height = max(leftHeight, rightHeight) + 1

    return height


def getDiameter(root):

    if (root == None):
    
        # Diameter of an empty tree will be 0.
        return 0
    

    # Get the height of left and right subtrees.
    leftHeight = getHeight(root.left)
    rightHeight = getHeight(root.right)

    # Recur for left subtree and get the diameter.
    leftDiameter = getDiameter(root.left)

    # Recur for right subtree and get the diameter.
    rightDiameter = getDiameter(root.right)

    # Diameter of given binary tree.
    diameter = max(leftDiameter, max(rightDiameter, leftHeight + rightHeight))

    return diameter


def diameterOfBinaryTree(root):

    return getDiameter(root)

Java (SE 1.8)

/*

    Time Complexity : O(N ^ 2)

    Space Complexity : O(N)



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

*/



public class Solution {



    // Returns the height of the tree.

    private static int getHeight(TreeNode root) {

        if (root == null) {

            // Height of empty tree is 0.

            return 0;

        }



        // Get the height of left subtree.

        int leftHeight = getHeight(root.left);



        // Get the height of right subtree.

        int rightHeight = getHeight(root.right);



        // Height of the given binary tree will be 1 greater than maximum of "leftHeight" and "rightHeight".

        int height = Math.max(leftHeight, rightHeight) + 1;



        return height;

    }



    // Returns the diameter of the given subtree rooted at the 'root' node.

    private static int getDiameter(TreeNode root) {

        if (root == null) {

            // Diameter of an empty tree will be 0.

            return 0;

        }



        // Get the height of left and right subtrees.

        int leftHeight = getHeight(root.left);

        int rightHeight = getHeight(root.right);



        // Recur for left subtree and get the diameter.

        int leftDiameter = getDiameter(root.left);



        // Recur for right subtree and get the diameter.

        int rightDiameter = getDiameter(root.right);



        // Diamter of given binary tree.

        int diameter = Math.max(leftDiameter, Math.max(rightDiameter, leftHeight + rightHeight));



        return diameter;

    }



    public static int diameterOfBinaryTree(TreeNode < Integer > root) {

        return getDiameter(root);

    }

}

Accepted Answer

Optimized Approach

The basic idea of this approach is to calculate the height of the subtree in the same recursion instead of calling getHeight() for it.

Let us define a recursive function getDiamter(TreeNode *root, int& height) which returns the diameter of the given subtree rooted at the “root” node. And, a variable “height” is passed by reference, which denotes the height of that subtree.

Algorithm

If the root node is NULL, assign “height” = 0 and return 0 because the height and diameter of an empty tree will be 0.
Initialize two variables, “leftHeight” and “rightHeight” to 0, which denotes the height of the left subtree and right subtree, respectively.
Recur for the left subtree and store the diameter of the left subtree in a variable i.e. leftDiameter = getDiameter(root->left, leftHeight)
Similarly, recur for the right subtree and store the diameter of the right subtree in a variable i.e. rightDiamater = getDiamter(root->right, rightHeight)
Update the height of the tree i.e. height = max(leftHeight, rightHeight) + 1
The diameter of the given tree will be the maximum of the following terms:
- “leftDiameter”
- “rightDiameter”
- “leftHeight” + “rightHeight”
Return the maximum of above terms i.e. max(leftDiameter, rightDiameter, leftHeight + rightHeight).

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 traversing all the tree nodes, which takes O(N) time, the overall time complexity will be O(N).

C++ (g++ 5.4)

/*

    Time Complexity : O(N)

    Space Complexity : O(N)

	

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

*/



int getDiameter(TreeNode *root, int &height)

{

    if (root == NULL)

    {

        // Height and diameter of an empty tree will be 0.

        height = 0;

        return 0;

    }



    // To store the height of left and right subtrees.

    int leftHeight = 0;

    int rightHeight = 0;



    // Recur for left subtree and get the height as well as diameter.

    int leftDiameter = getDiameter(root->left, leftHeight);



    // Recur for right subtree and get the height as well as diameter.

    int rightDiameter = getDiameter(root->right, rightHeight);



    // Update the height of the given binary tree.

    height = max(leftHeight, rightHeight) + 1;



    // Diameter of given binary tree.

    int diameter = max(leftDiameter, max(rightDiameter, leftHeight + rightHeight));



    return diameter;

}



int diameterOfBinaryTree(TreeNode *root)

{

    // Initialize a variable to store the height of the of binary tree.

    int height = 0;



    // Recursive function to find diameter.

    return getDiameter(root, height);

}

Java (SE 1.8)

/*

    Time Complexity : O(N)

    Space Complexity : O(N)



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

*/



public class Solution {



    // Returns the diameter of the given subtree rooted at the 'root' node.

    private static int getDiameter(TreeNode root, int[] height) {

        if (root == null) {

            // Height and diameter of an empty tree will be 0.

            height[0] = 0;

            return 0;

        }



        // To store the height of left and right subtrees.

        int[] leftHeight = { 0 };

        int[] rightHeight = { 0 };



        // Recur for left subtree and get the height as well as diameter.

        int leftDiameter = getDiameter(root.left, leftHeight);



        // Recur for right subtree and get the height as well as diameter.

        int rightDiameter = getDiameter(root.right, rightHeight);



        // Update the height of the given binary tree.

        height[0] = Math.max(leftHeight[0], rightHeight[0]) + 1;



        // Diameter of given binary tree.

        int diameter = Math.max(leftDiameter, Math.max(rightDiameter, leftHeight[0] + rightHeight[0]));



        return diameter;

    }



    public static int diameterOfBinaryTree(TreeNode root) {

        // To store the height of the of binary tree.

        int[] height = { 0 };



        return getDiameter(root, height);

    }

}

Python (3.5)

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

def getDiameter(root, height):

    if (root == None):
    
        # Height and diameter of an empty tree will be 0.
        height[0] = 0
        return 0
    

    # To store the height of left and right subtrees.
    leftHeight = [0]
    rightHeight = [0]

    # Recur for left subtree and get the height as well as diameter.
    leftDiameter = getDiameter(root.left, leftHeight)

    # Recur for right subtree and get the height as well as diameter.
    rightDiameter = getDiameter(root.right, rightHeight)

    # Update the height of the given binary tree.
    height[0] = max(leftHeight[0], rightHeight[0]) + 1

    # Diameter of given binary tree.
    diameter = max(leftDiameter, max(rightDiameter, leftHeight[0] + rightHeight[0]))

    return diameter


def diameterOfBinaryTree(root):

    # Initialize a variable to store the height of the of binary tree.
    height = [0]

    # Recursive function to find diameter.
    return getDiameter(root, height)

You are given a Binary Tree. You are supposed to return the length of the diameter of the tree.

The diameter of a binary tree is the length of the longest path between any two end nodes in a tree.

Note :

For Example :

Input Format :

Explanation :

Note :

Output Format :

Note :

Constraints :

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