Last Updated: 27 Oct, 2021

Longest ZigZag Path In Binary Tree

Moderate

Asked in company

Problem statement

You have been practicing on the topic binary tree for a few days. Your friend challenged you by giving you a binary tree and asked you to find the longest ZigZag path in the binary tree. ZigZag path of the binary tree is defined as below:

1. Choose any node in the binary tree and a direction to move further.
2. If the current direction is left, move to the right child of the current node, otherwise move to the left node.
3. Change the direction from right to left or from left to right.
4. Repeat the above steps until you are unable to move.
The longest path is defined as the number of nodes visited - 1.

Note :

1. Two nodes may have the same value associated with it.
2. The root node will be fixed and will be provided in the function.

Input Format :

The first line of the input contains a single integer 'T', representing the number of test cases.

The first line of each test case contains an integer 'N', which denotes the number of nodes in the tree.

The second line of each test case will contain the values of the nodes of the tree in the level order form ( -1 for 'NULL' node). Refer to the example for further clarification.

Example :

Consider the binary tree

The input of the tree depicted in the image above will be like : 

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).

Output Format :

For each test case, print the longest ZigZag path in the binary tree.

Print the output of each test case 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 <= 10
1 <= N <= 10^5

It is guaranteed that the sum of ‘N’ over all test cases doesn’t exceed 10^5.

Time Limit : 1 sec

Solve now

Approaches

01 Approach

The basic idea is to store the previous direction taken for each node. We use bfs to find the length of the longest ZigZag path. We use a queue to traverse the tree and an unordered map to store the previous direction and length of the longest path till the current node. We will represent the left direction with “-1”, whereas the right direction with ”1”.

Here is the algorithm :

Create a variable (say, ‘ANS’) to store the longest path.
Create an unordered map (say, ’MP’) that will store the direction and length of the path.
Create a queue (say, ‘Q’) to traverse the tree.
Check if the left child of ‘ROOT’ exists,
- Push left child of ‘ROOT’ in ‘Q’.
- Store the direction of the left child of ‘ROOT’ with “-1” and length 1.
Similarly, check if the right child of ‘ROOT’ exists,
- Push right child of ‘ROOT’ in ‘Q’.
- Store the direction of the right child of ‘ROOT’ with “1” and length 1.
Run a loop till ‘Q’ is not empty.
- Create a variable (say, ‘CURR’) to store the ‘Q’’s current node.
- Pop the element from ‘Q’.
- Update ‘ANS’ by a maximum of ‘ANS’ and length till ‘CURR’ node.
- Check if the previous direction taken was left.
  - Check if the right child of ‘CURR’ exists,
    - Push right child of ‘CURR’ in ‘Q’.
    - Store the direction of the right child of ‘ROOT’ with “1” and length 1 + length until the ‘CURR’ node.
  - Check if the left child of ‘CURR’ exists,
    - Push left child of ‘CURR’ in ‘Q’.
    - Store the direction of the left child of ‘ROOT’ with “-1” and length 1.
- Similarly, check if the previous direction taken was right.
  - Check if the right child of ‘CURR’ exists,
    - Push right child of ‘CURR’ in ‘Q’.
    - Store the direction of the right child of ‘ROOT’ with “1” and length 1.
  - Check if the left child of ‘CURR’ exists,
    - Push left child of ‘CURR’ in ‘Q’.
    - Store the direction of the left child of ‘ROOT’ with “-1” and length 1 + length till the ‘CURR’ node.
Finally, return ‘ANS’.

Solve now

02 Approach

The basic idea is to store the direction of the previous node for a current node. We use DFS to traverse the tree while checking for the previous direction traversed. If the previous direction traversed was left, we add the result of the longest ZigZag path of the right subtree, else if the previous direction traversed was right, we add the result of the longest ZigZag path of the left subtree. To find the longest path, we take the maximum of both of these and return the result.

Here is the algorithm :

Create a variable (say, ‘ANS’) that will be used to store the longest path and initialize it with 0.
Call the DFS function to find the result.
Finally, return ‘ANS’.

DFS(‘ROOT’, ‘DIRECTION’, ‘ANS’) (where ‘ROOT’ is the root node of the tree, ‘DIRECTION’ is the previous direction taken, and ‘ANS’ stores the longest path)

Base case:
- Check if ‘ROOT’ is equal to NULL, return 0.
Call the DFS function recursively on the left subtree and update the direction to ‘L’ and store the value returned by the function in a variable (say, ‘LEFT’).
Similarly, call the DFS function recursively on the right subtree and update the direction to ‘R’ and store the value returned by the function in a variable (say, ‘RIGHT’).
Update the ‘ANS’ by a maximum of ‘LEFT’, ‘RIGHT’, and ‘ANS’ itself.
Check if ‘DIRECTION’ is equal to ‘R’.
- Return 1 + ‘LEFT.
Else,
- Return 1 + ‘RIGHT’.

Solve now