Max Increasing Path

Moderate
0/80
Average time to solve is 15m
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Citrix

Problem statement

You are given a binary tree with ‘N’ number of nodes. Your task is to find the length of the longest path of nodes having values in increasing order. This path need not start and end at the root and leaf node respectively. That is, it can start and end from any node.

A binary tree is a hierarchical data structure in which each node has at most two children.

Note: A Path should be linear, i.e it should start from a node and ends in a subtree of that node only.

Example:

For the given tree, the maximum consecutive path is 10 -> 11 -> 12, whose length is 3.

Hence, the output will be 3.

Detailed explanation ( Input/output format, Notes, Images )
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 follow.
The first line of each test case contains the elements of the tree in the level order form separated by a single space.
If any node does not have a left or right child, take -1 in its place.

Example:

Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 in its place.

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, return the length of the longest path of nodes having values in increasing order.

The output of each test case should be printed in a separate line.
Note: 
You are not required to print anything, it has already been taken care of. Just implement the function.
Constraints : 
1 <= T <= 100
1 <= N <= 3000
0 <= nodeVal<= 10^5 

Where, nodeVal denotes the value associated with the node.

Time Limit : 1 sec.
Sample Input 1:
2
10 9 4 -1 -1 5 8 -1 6 -1 -1 3 -1 -1 -1
11 7 3 9 4 -1 -1 -1 -1 -1 -1
Sample Output 1 :
3
2
Explanation For Sample Output 1 :
Test Case 1:

For the given level order traversal the tree will be:-


For this tree, the maximum consecutive path will be 4 -> 5 -> 6, whose length is 3. Hence the output will be 3.

Test Case 2:

For the given level order traversal the tree will be:-


For the given tree the maximum consecutive path will be 7 -> 9, whose length is 2. Hence the output will be 2.
Sample Input 2:
2
10 11 9 13 12 13 8 -1 -1 -1 -1 -1 -1 -1 -1
5 8 11 9 -1 -1 10 6 -1 15 -1 -1 -1 -1 -1
Sample Output 2 :
3
3
Hint

Think of recursively finding the path length for the left and right subtree.

Approaches (2)
Depth First Search

Every node in the Binary Tree can either become part of the path which is starting from one of its parent nodes or a new path can start from this node itself. The key idea is to recursively find the path length for the left and right subtree and then return the maximum. 

 

Algorithm :
 

  1. Two functions namely, dfs1 and df2 will be used in this with the root node as the parameter passed to them.
  2. In dfs1 :

        1. If the root node is NULL or it does not exist, simply return.

        2. Initialize three variables, say, ‘LEN’ as 1, ‘X’ as 0 and ‘Y’ as 0.

        3. If the left child of the root node exists and it is greater than the root node, then call function ‘dfs2’ with left child as the parameter and store the result in variable ‘X’.

        4. If the right child of the root node exists and it is greater than the root node, then call function ‘dfs2’ with right child as the parameter and store the result in variable ‘Y’.

        5. Return ( 1 + max(‘X’, ‘Y’) ).

 

3.     In dfs2 :

        1. If the root node is NULL or it does not exist, simply return.

        2. Initialize three variables, say, ‘L’ as 0, ‘R’ as 0 and ‘X’ as 0.

        3. Call function ‘dfs2’ with parameter as root node.

        4. If the left child of the root node exists then call function ‘dfs1’ with left child as the parameter and store the result in variable ‘L’.

        5. If the right child of the root node exists then call function ‘dfs1’ with right child as the parameter and store the result in variable ‘R’.

        6. Return max( ‘L’, max(‘R’, ‘X’) ).

 

4.    In function ‘longestIncreasingPath’, call function ‘dfs1’ with parameter as root node and return it.

Time Complexity

O(N^2), where ‘N’ is the number of nodes in a Binary Tree.

 

Since, there are N nodes in total, and each node is traversed N number of times, hence time complexity will be O(N^2).

Space Complexity

O(‘N’), where ‘N’ is the number of nodes in a Binary Tree.

 

Space complexity will be equal to the height of the given binary tree due to the stack size of recursion, and in the worst case, the height of the tree can be N. So the space complexity will be O(N).

Code Solution
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Max Increasing Path
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