Longest Univalue Path

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Problem statement

You are given a binary tree, the task is to find out the length of the longest path which contains nodes with the exact same value. It is not necessary for the path to pass through the root of the binary tree.

Between two nodes, the length of the path can be defined as the number of edges contained between them.

For example, consider the following binary tree:

            7
           / \
          7   7
         / \   \
        8  3    7

For the above tree, the length of the longest path where each node in the path has the same value is 3 and path is 7 -> 7 -> 7 -> 7.

Detailed explanation ( Input/output format, Notes, Images )
Input Format:
The first line of input contains an integer ‘T’ representing the number of test cases. Then the test cases follow.

The only line of each test case 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 on its place.

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, a single integer denoting the length of the longest path where each node in the path has the same value is printed.

The output for each test case is to be printed on 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
1 <= N <= 3000
-10^9 <= data <= 10^9 and data != -1

Where ‘T’ is the number of test cases, ‘N’ is the total number of nodes in the binary tree, and “data” is the value of the binary tree node.

Time Limit: 1sec
Sample Input 1:
2
7 7 7 1 1 -1 7 -1 -1 -1 -1 -1 -1
10 3 4 3 3 -1 5 -1 -1 -1 -1 -1 -1
Sample Output 1:
3
2
Explanation of Sample Input 1:
            7
           / \
          7   7
         / \   \
        1  1    7

For the first test case, the length of the longest path where each node in the path has the same value is 3 and path is 7 -> 7 -> 7 -> 7.

            10
           / \
          3   4
         / \   \
        3  3    5

For the second test case, the length of the longest path where each node in the path has the same value is 2 and path is 3 -> 3 -> 3.
Sample Input 2:
2
1 4 5 4 4 -1 5 -1 -1 -1 -1 -1 -1
5 4 5 1 1 -1 5 -1 -1 -1 -1 -1 -1
Sample Output 2:
2
2
Hint

Recursively traverse the binary tree and get the maximum path extending from this current node.

Approaches (1)
Recursive Approach

The idea is to use recursion to traverse the given binary tree and update the maximum path extending from the current node by the maximum of left and right path (only if their data is same as the root's data).

 

Let length(root) be the length of the longest path that extends from the root. That will be 1 + length(root.left) if root.left exists and has the same value as root. Similarly for the root.right case.

 

While we are computing lengths, each possible answer will be the sum of the lengths in both directions from that root. We record these possible answers and return the best one.

 

Algorithm:

  1. Check for the base case. If the root is null, return 0.
  2. Recursively call to the left and the right subtrees and store their results in leftMax and in rightMax variables respectively.
  3. Check if the root and it’s left child has the same value or not. If yes, increase the leftMax variable by 1.
  4. Check if the root and it’s right child has the same value or not. If yes, increase the rightMax variable by 1.
  5. Update the global max(the longest path among every node path) with max( max, leftMax + rightMax).
  6. We then return max (leftMax, rightMax) to the parent’s node as the longest path so far (only through one side is a valid path).
Time Complexity

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

 

Since in the worst case (Skewed Trees), we might be visiting all the nodes of the binary tree, which is ‘N’ in the worst case.

So, the overall time complexity will be linear, i.e. O(N).

Space Complexity

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

 

We are using O(H) extra space for the call stack where ‘H’ is the height of the tree, and height of a skewed binary tree could be ‘N’ in the worst case.

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