Number of turns to reach from one node to another

Easy
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Average time to solve is 15m
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Problem statement

Given a binary tree and two node values, the first one is ‘source’ node value and the second one is ‘destination’ node value. Your task is to find a number of turns to reach from ‘source’ node to the ‘destination’ node.

A turn is defined as rotation in the path from ‘source’ to ‘destination’

For example -

The path from node 1( ‘source’ as blue) to -9 ( ‘destination’ as blue) is shown in the diagram.

example

All the nodes that come in the path are represented by orange colour.

According to the above path, we take a turn at node 4, node 10 and at node 3 to reach the destination node i.e. -9.

Hence the total number of turns required to reach at ‘1’ to ‘-9’ is ‘3’, return the total number of turns 3.

Note: 
1. You are not required to print the output explicitly, it has already been taken care of. Just implement the function and return the number of turns required to reach from one ‘source’ node to ‘destination’ node.
2. It is sure that every node value will present exactly once in the binary tree.
Detailed explanation ( Input/output format, Notes, Images )
Input Format: 
The first line of input contains an integer ‘T’ denoting the number of test cases.
The next ‘2*T’ lines represent the ‘T’ test cases.

The first line of input contains two space-separated integers, the first denotes the given ‘source’ node value and the second denotes the ‘destination’ node value.

The second line of input 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. Refer to the example below.

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 :

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).
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 every test case, return the number of turns.
Constraint : 
1 <= T <= 100
2 <= N <= 3000 
-10^9 <= S, D <= 10^9
-10^9 <= data <= 10^9

Where ‘T’ represents the number of test cases, ‘N’ is the number of nodes in the tree, ‘S’ and ‘D’ denotes the value of ‘source’ node and ‘destination’ node, and ‘data’ denotes data contained in the node of a binary tree.

Time Limit: 1 sec
Sample Input 1:
2
4 6
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
3 6
1 2 4 3 -1 5 -1 -1 -1 -1 6 -1 -1
Sample output 1:
1
3
Explanation of sample input 1:
 Test case 1:

Sampletest1

According to the above path, we take a turn at node 1 to reach the destination node at 6 from ‘source’ node ‘4’. 
Hence the total number of turns required to reach at ‘4’ to ‘6’ is ‘1’, return the total number of turns 1.

Test case 2:

Sampletest2

According to the above path, we take a turn at node 1, at node 2, and at node 5 to reach the destination node at ‘6’ from ‘source’ node ‘3’. 
Hence the total number of turns required to reach at ‘4’ to ‘6’ is ‘3’, return the total number of turns 3.
Sample Input 2 :
2
7 3
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
4 3
1 -1 2 -1 3 -1 4 -1 5 -1 -1
Sample Output 2:
2
0
Hint

Think recursively.

Approaches (2)
LCA Based Recursion

The idea is to first find the LCA(source node and destination node). From LCA to find the number of turns to reach the ‘source’ node as well as ‘destination’ node.

 

Find the LCA of ‘source’ node and ‘destination’ node.

  • Use a recursive function lowestCommonAncestor( root, source, destination) to find LCA.
  • ‘leftAns’ = lowestCommonAncestor( root->left, source, destination)
    • If our answer is present on the left side of ‘root’.
  • ‘rightAns’ = lowestCommonAncestor( root->right, source, destination)
    • If our answer is present on the right side of ‘root’.
  • Check if ‘rightAns’ and ‘leftAns’ are present then the root is LCA.

Find the number of turns from LCA to ‘source’ node.

  • Call the countHelper(root, targetNode, turn)
  • The root is the current node, ‘targetNode’ can be either ‘source’ node or ‘destination’ node. turns value represents last call subtree if(root->left) then turn will be ‘True’ else it will be ‘False’.

Find the number of turns from LCA to ‘destination’ node.

  • Find in a similar way.
  • If the total number of turns required to reach ‘source’ is ‘x’ and to reach ‘destination’ is ‘y’ then return ‘x+y+1’.

  

Time Complexity

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

 

We are exploring every node at max twice.

Space Complexity

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

 

We are using an ‘N’ size of the call stack.

Code Solution
(100% EXP penalty)
Number of turns to reach from one node to another
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