Diameter Of Binary Tree

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Problem statement

You are given a Binary Tree.


Return the length of the diameter of the tree.


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

The number of edges between two nodes represents the length of the path between them.
Example :
Input: Consider the given binary tree:

Example

Output: 6

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


Detailed explanation ( Input/output format, Notes, Images )
Input Format:
The only line contains elements 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 below image will be:

alt text

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).

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:
Return a single integer i.e. the diameter of the tree.


Note :
You do not need to print anything; it has already been taken care of. Just implement the given function.
Sample Input 1 :
1 2 3 4 7 -1 -1 -1 -1 -1 -1


Sample Output 1 :
3


Explanation of Sample Input 1 :
The binary tree in the given test case will look like

The length of the diameter of the above tree is the length of the path between nodes 4 and 3, i.e., 4->2->1->3. Hence the output will be 3.


Sample Input 2 :
1 2 3 -1 -1 -1 -1


Sample Output 2 :
2


Explanation of Sample Input 2 :
The binary tree in the given test case will look like

The length of the diameter of the above tree is the length of the path between nodes 2 and 3, i.e., 2.


Expected Time Complexity:
Try to do this in O(n).


Constraints:
1 <= n <= 10000

Where 'n' is the total number of nodes in the binary tree.

Time Limit: 1 sec
Hint

Try to divide your problem into smaller sub-problems using recursion.

Approaches (2)
Recursion

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

Now, there are three possible cases:

 

  1. The diameter of the tree is present in the left subtree.
  2. The diameter of the tree is present in the right subtree.
  3. 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).
Time Complexity

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).


 

Space Complexity

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).

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Diameter Of Binary Tree
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