# Number of turns to reach from one node to other in binary tree

Given a binary tree and two nodes. The task is to count the number of turns needs to reach from one node to another node of the Binary tree.

Examples:

```Input:   Below Binary Tree and two nodes
5 & 6
1
/
2        3
/       /
4     5   6     7
/         /
8         9   10
Output: Number of Turns needed to reach
from 5 to 6:  3

Input: For above tree if two nodes are 1 & 4
Output: Straight line : 0 turn
```

Idea based on the Lowest Common Ancestor in a Binary Tree
We have to follow the step.
1…Find the LCA of given two node
2…Given node present either on the left side or right side or equal to LCA.
…. According to above condition over program falls under two Case.

```Case 1:
If none of the nodes is equal to
LCA, we get these nodes either on
the left side or right side.
We call two functions for each node.
....a)  if (CountTurn(LCA->right, first,
false, &Count)
|| CountTurn(LCA->left, first,
true, &Count)) ;
....b)  Same for second node.
....Here Count is used to store number of
turns need to reached the target node.

Case 2:
If one of the nodes is equal to LCA_Node.
Then we count only number of turns needs
to reached the second node.
If LCA == (Either first or second)
....a)  if (countTurn(LCA->right, second/first,
false, &Count)
|| countTurn(LCA->left, second/first,
true, &Count)) ;
```

3… Working of CountTurn Function
// we pass turn true if we move
// left subtree and false if we
// move right subTree

```CountTurn(LCA, Target_node, count, Turn)

// if found the key value in tree
if (root->key == key)
return true;
case 1:
If Turn is true that means we are
in left_subtree
If we going left_subtree then there
is no need to increment count
else
Increment count and set turn as false
case 2:
if Turn is false that means we are in
right_subtree
if we going right_subtree then there is
no need to increment count else
increment count and set turn as true.

// if key is not found.
return false;
```

Below the implementation of above idea.

## C++

 `// C++ Program to count number of turns ` `// in a Binary Tree. ` `#include ` `using` `namespace` `std; ` ` `  `// A Binary Tree Node ` `struct` `Node { ` `    ``struct` `Node* left, *right; ` `    ``int` `key; ` `}; ` ` `  `// Utility function to create a new  ` `// tree Node ` `Node* newNode(``int` `key) ` `{ ` `    ``Node* temp = ``new` `Node; ` `    ``temp->key = key; ` `    ``temp->left = temp->right = NULL; ` `    ``return` `temp; ` `} ` ` `  `// Utility function to find the LCA of  ` `// two given values n1 and n2. ` `struct` `Node* findLCA(``struct` `Node* root,  ` `                        ``int` `n1, ``int` `n2) ` `{ ` `    ``// Base case ` `    ``if` `(root == NULL) ` `        ``return` `NULL; ` ` `  `    ``// If either n1 or n2 matches with  ` `    ``// root's key, report the presence by  ` `    ``// returning root (Note that if a key  ` `    ``// is ancestor of other, then the ` `    ``// ancestor key becomes LCA ` `    ``if` `(root->key == n1 || root->key == n2) ` `        ``return` `root; ` ` `  `    ``// Look for keys in left and right subtrees ` `    ``Node* left_lca = findLCA(root->left, n1, n2); ` `    ``Node* right_lca = findLCA(root->right, n1, n2); ` ` `  `    ``// If both of the above calls return  ` `    ``// Non-NULL, then one key is present  ` `    ``// in once subtree and other is present  ` `    ``// in other, So this node is the LCA ` `    ``if` `(left_lca && right_lca) ` `        ``return` `root; ` ` `  `    ``// Otherwise check if left subtree or right ` `    ``// subtree is LCA ` `    ``return` `(left_lca != NULL) ? left_lca :  ` `                                ``right_lca; ` `} ` ` `  `// function count number of turn need to reach ` `// given node from it's LCA we have two way to ` `bool` `CountTurn(Node* root, ``int` `key, ``bool` `turn, ` `                                   ``int``* count) ` `{ ` `    ``if` `(root == NULL) ` `        ``return` `false``; ` ` `  `    ``// if found the key value in tree ` `    ``if` `(root->key == key) ` `        ``return` `true``; ` ` `  `    ``// Case 1: ` `    ``if` `(turn == ``true``) { ` `        ``if` `(CountTurn(root->left, key, turn, count)) ` `            ``return` `true``; ` `        ``if` `(CountTurn(root->right, key, !turn, count)) { ` `            ``*count += 1; ` `            ``return` `true``; ` `        ``} ` `    ``}  ` `    ``else` `// Case 2: ` `    ``{ ` `        ``if` `(CountTurn(root->right, key, turn, count)) ` `            ``return` `true``; ` `        ``if` `(CountTurn(root->left, key, !turn, count)) { ` `            ``*count += 1; ` `            ``return` `true``; ` `        ``} ` `    ``} ` `    ``return` `false``; ` `} ` ` `  `// Function to find nodes common to given two nodes ` `int` `NumberOFTurn(``struct` `Node* root, ``int` `first,  ` `                                   ``int` `second) ` `{ ` `    ``struct` `Node* LCA = findLCA(root, first, second); ` ` `  `    ``// there is no path between these two node ` `    ``if` `(LCA == NULL) ` `        ``return` `-1; ` `    ``int` `Count = 0; ` ` `  `    ``// case 1: ` `    ``if` `(LCA->key != first && LCA->key != second) { ` `         `  `        ``// count number of turns needs to reached ` `        ``// the second node from LCA ` `        ``if` `(CountTurn(LCA->right, second, ``false``, ` `                                           ``&Count) ` `            ``|| CountTurn(LCA->left, second, ``true``,  ` `                                           ``&Count)) ` `            ``; ` `         `  `        ``// count number of turns needs to reached  ` `        ``// the first node from LCA ` `        ``if` `(CountTurn(LCA->left, first, ``true``,  ` `                                       ``&Count) ` `            ``|| CountTurn(LCA->right, first, ``false``,  ` `                                       ``&Count)) ` `            ``; ` `        ``return` `Count + 1; ` `    ``} ` ` `  `    ``// case 2: ` `    ``if` `(LCA->key == first) { ` ` `  `        ``// count number of turns needs to reached  ` `        ``// the second node from LCA ` `        ``CountTurn(LCA->right, second, ``false``, &Count); ` `        ``CountTurn(LCA->left, second, ``true``, &Count); ` `        ``return` `Count; ` `    ``} ``else` `{ ` ` `  `        ``// count number of turns needs to reached ` `        ``// the first node from LCA1 ` `        ``CountTurn(LCA->right, first, ``false``, &Count); ` `        ``CountTurn(LCA->left, first, ``true``, &Count); ` `        ``return` `Count; ` `    ``} ` `} ` ` `  `// Driver program to test above functions ` `int` `main() ` `{ ` `    ``// Let us create binary tree given in the above ` `    ``// example ` `    ``Node* root = newNode(1); ` `    ``root->left = newNode(2); ` `    ``root->right = newNode(3); ` `    ``root->left->left = newNode(4); ` `    ``root->left->right = newNode(5); ` `    ``root->right->left = newNode(6); ` `    ``root->right->right = newNode(7); ` `    ``root->left->left->left = newNode(8); ` `    ``root->right->left->left = newNode(9); ` `    ``root->right->left->right = newNode(10); ` ` `  `    ``int` `turn = 0; ` `    ``if` `((turn = NumberOFTurn(root, 5, 10))) ` `        ``cout << turn << endl; ` `    ``else` `        ``cout << ``"Not Possible"` `<< endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `//A Java Program to count number of turns ` `//in a Binary Tree. ` `public` `class` `Turns_to_reach_another_node { ` ` `  `    ``// making Count global such that it can get  ` `    ``// modified by different methods ` `    ``static` `int` `Count; ` ` `  `    ``// A Binary Tree Node ` `    ``static` `class` `Node { ` `        ``Node left, right; ` `        ``int` `key; ` ` `  `        ``// Constructor ` `        ``Node(``int` `key) { ` `            ``this``.key = key; ` `            ``left = ``null``; ` `            ``right = ``null``; ` `        ``} ` `    ``} ` ` `  `    ``// Utility function to find the LCA of ` `    ``// two given values n1 and n2. ` `    ``static` `Node findLCA(Node root, ``int` `n1, ``int` `n2) { ` `        ``// Base case ` `        ``if` `(root == ``null``) ` `            ``return` `null``; ` ` `  `        ``// If either n1 or n2 matches with ` `        ``// root's key, report the presence by ` `        ``// returning root (Note that if a key ` `        ``// is ancestor of other, then the ` `        ``// ancestor key becomes LCA ` `        ``if` `(root.key == n1 || root.key == n2) ` `            ``return` `root; ` ` `  `        ``// Look for keys in left and right subtrees ` `        ``Node left_lca = findLCA(root.left, n1, n2); ` `        ``Node right_lca = findLCA(root.right, n1, n2); ` ` `  `        ``// If both of the above calls return ` `        ``// Non-NULL, then one key is present ` `        ``// in once subtree and other is present ` `        ``// in other, So this node is the LCA ` `        ``if` `(left_lca != ``null` `&& right_lca != ``null``) ` `            ``return` `root; ` ` `  `        ``// Otherwise check if left subtree or right ` `        ``// subtree is LCA ` `        ``return` `(left_lca != ``null``) ? left_lca : right_lca; ` `    ``} ` ` `  `    ``// function count number of turn need to reach ` `    ``// given node from it's LCA we have two way to ` `    ``static` `boolean` `CountTurn(Node root, ``int` `key, ``boolean` `turn) { ` `        ``if` `(root == ``null``) ` `            ``return` `false``; ` ` `  `        ``// if found the key value in tree ` `        ``if` `(root.key == key) ` `            ``return` `true``; ` ` `  `        ``// Case 1: ` `        ``if` `(turn == ``true``) { ` `            ``if` `(CountTurn(root.left, key, turn)) ` `                ``return` `true``; ` `            ``if` `(CountTurn(root.right, key, !turn)) { ` `                ``Count += ``1``; ` `                ``return` `true``; ` `            ``} ` `        ``} ``else` `// Case 2: ` `        ``{ ` `            ``if` `(CountTurn(root.right, key, turn)) ` `                ``return` `true``; ` `            ``if` `(CountTurn(root.left, key, !turn)) { ` `                ``Count += ``1``; ` `                ``return` `true``; ` `            ``} ` `        ``} ` `        ``return` `false``; ` `    ``} ` ` `  `    ``// Function to find nodes common to given two nodes ` `    ``static` `int` `NumberOfTurn(Node root, ``int` `first, ``int` `second) { ` `        ``Node LCA = findLCA(root, first, second); ` ` `  `        ``// there is no path between these two node ` `        ``if` `(LCA == ``null``) ` `            ``return` `-``1``; ` `        ``Count = ``0``; ` ` `  `        ``// case 1: ` `        ``if` `(LCA.key != first && LCA.key != second) { ` ` `  `            ``// count number of turns needs to reached ` `            ``// the second node from LCA ` `            ``if` `(CountTurn(LCA.right, second, ``false``) ` `                    ``|| CountTurn(LCA.left, second, ``true``)) ` `                ``; ` ` `  `            ``// count number of turns needs to reached ` `            ``// the first node from LCA ` `            ``if` `(CountTurn(LCA.left, first, ``true``) ` `                    ``|| CountTurn(LCA.right, first, ``false``)) ` `                ``; ` `            ``return` `Count + ``1``; ` `        ``} ` ` `  `        ``// case 2: ` `        ``if` `(LCA.key == first) { ` ` `  `            ``// count number of turns needs to reached ` `            ``// the second node from LCA ` `            ``CountTurn(LCA.right, second, ``false``); ` `            ``CountTurn(LCA.left, second, ``true``); ` `            ``return` `Count; ` `        ``} ``else` `{ ` ` `  `            ``// count number of turns needs to reached ` `            ``// the first node from LCA1 ` `            ``CountTurn(LCA.right, first, ``false``); ` `            ``CountTurn(LCA.left, first, ``true``); ` `            ``return` `Count; ` `        ``} ` `    ``} ` ` `  `    ``// Driver program to test above functions ` `    ``public` `static` `void` `main(String[] args) { ` `        ``// Let us create binary tree given in the above ` `        ``// example ` `        ``Node root = ``new` `Node(``1``); ` `        ``root.left = ``new` `Node(``2``); ` `        ``root.right = ``new` `Node(``3``); ` `        ``root.left.left = ``new` `Node(``4``); ` `        ``root.left.right = ``new` `Node(``5``); ` `        ``root.right.left = ``new` `Node(``6``); ` `        ``root.right.right = ``new` `Node(``7``); ` `        ``root.left.left.left = ``new` `Node(``8``); ` `        ``root.right.left.left = ``new` `Node(``9``); ` `        ``root.right.left.right = ``new` `Node(``10``); ` ` `  `        ``int` `turn = ``0``; ` `        ``if` `((turn = NumberOfTurn(root, ``5``, ``10``)) != ``0``) ` `            ``System.out.println(turn); ` `        ``else` `            ``System.out.println(``"Not Possible"``); ` `    ``} ` ` `  `} ` `// This code is contributed by Sumit Ghosh `

## C#

 `using` `System; ` ` `  `//A C# Program to count number of turns ` `//in a Binary Tree. ` `public` `class` `Turns_to_reach_another_node ` `{ ` ` `  `    ``// making Count global such that it can get  ` `    ``// modified by different methods ` `    ``public` `static` `int` `Count; ` ` `  `    ``// A Binary Tree Node ` `    ``public` `class` `Node ` `    ``{ ` `        ``public` `Node left, right; ` `        ``public` `int` `key; ` ` `  `        ``// Constructor ` `        ``public` `Node(``int` `key) ` `        ``{ ` `            ``this``.key = key; ` `            ``left = ``null``; ` `            ``right = ``null``; ` `        ``} ` `    ``} ` ` `  `    ``// Utility function to find the LCA of ` `    ``// two given values n1 and n2. ` `    ``public` `static` `Node findLCA(Node root, ``int` `n1, ``int` `n2) ` `    ``{ ` `        ``// Base case ` `        ``if` `(root == ``null``) ` `        ``{ ` `            ``return` `null``; ` `        ``} ` ` `  `        ``// If either n1 or n2 matches with ` `        ``// root's key, report the presence by ` `        ``// returning root (Note that if a key ` `        ``// is ancestor of other, then the ` `        ``// ancestor key becomes LCA ` `        ``if` `(root.key == n1 || root.key == n2) ` `        ``{ ` `            ``return` `root; ` `        ``} ` ` `  `        ``// Look for keys in left and right subtrees ` `        ``Node left_lca = findLCA(root.left, n1, n2); ` `        ``Node right_lca = findLCA(root.right, n1, n2); ` ` `  `        ``// If both of the above calls return ` `        ``// Non-NULL, then one key is present ` `        ``// in once subtree and other is present ` `        ``// in other, So this node is the LCA ` `        ``if` `(left_lca != ``null` `&& right_lca != ``null``) ` `        ``{ ` `            ``return` `root; ` `        ``} ` ` `  `        ``// Otherwise check if left subtree or right ` `        ``// subtree is LCA ` `        ``return` `(left_lca != ``null``) ? left_lca : right_lca; ` `    ``} ` ` `  `    ``// function count number of turn need to reach ` `    ``// given node from it's LCA we have two way to ` `    ``public` `static` `bool` `CountTurn(Node root, ``int` `key, ``bool` `turn) ` `    ``{ ` `        ``if` `(root == ``null``) ` `        ``{ ` `            ``return` `false``; ` `        ``} ` ` `  `        ``// if found the key value in tree ` `        ``if` `(root.key == key) ` `        ``{ ` `            ``return` `true``; ` `        ``} ` ` `  `        ``// Case 1: ` `        ``if` `(turn == ``true``) ` `        ``{ ` `            ``if` `(CountTurn(root.left, key, turn)) ` `            ``{ ` `                ``return` `true``; ` `            ``} ` `            ``if` `(CountTurn(root.right, key, !turn)) ` `            ``{ ` `                ``Count += 1; ` `                ``return` `true``; ` `            ``} ` `        ``} ` `        ``else` `// Case 2: ` `        ``{ ` `            ``if` `(CountTurn(root.right, key, turn)) ` `            ``{ ` `                ``return` `true``; ` `            ``} ` `            ``if` `(CountTurn(root.left, key, !turn)) ` `            ``{ ` `                ``Count += 1; ` `                ``return` `true``; ` `            ``} ` `        ``} ` `        ``return` `false``; ` `    ``} ` ` `  `    ``// Function to find nodes common to given two nodes ` `    ``public` `static` `int` `NumberOfTurn(Node root, ``int` `first, ``int` `second) ` `    ``{ ` `        ``Node LCA = findLCA(root, first, second); ` ` `  `        ``// there is no path between these two node ` `        ``if` `(LCA == ``null``) ` `        ``{ ` `            ``return` `-1; ` `        ``} ` `        ``Count = 0; ` ` `  `        ``// case 1: ` `        ``if` `(LCA.key != first && LCA.key != second) ` `        ``{ ` ` `  `            ``// count number of turns needs to reached ` `            ``// the second node from LCA ` `            ``if` `(CountTurn(LCA.right, second, ``false``)  ` `                ``|| CountTurn(LCA.left, second, ``true``)) ` `            ``{ ` `                ``; ` `            ``} ` ` `  `            ``// count number of turns needs to reached ` `            ``// the first node from LCA ` `            ``if` `(CountTurn(LCA.left, first, ``true``)  ` `                ``|| CountTurn(LCA.right, first, ``false``)) ` `            ``{ ` `                ``; ` `            ``} ` `            ``return` `Count + 1; ` `        ``} ` ` `  `        ``// case 2: ` `        ``if` `(LCA.key == first) ` `        ``{ ` ` `  `            ``// count number of turns needs to reached ` `            ``// the second node from LCA ` `            ``CountTurn(LCA.right, second, ``false``); ` `            ``CountTurn(LCA.left, second, ``true``); ` `            ``return` `Count; ` `        ``} ` `        ``else` `        ``{ ` ` `  `            ``// count number of turns needs to reached ` `            ``// the first node from LCA1 ` `            ``CountTurn(LCA.right, first, ``false``); ` `            ``CountTurn(LCA.left, first, ``true``); ` `            ``return` `Count; ` `        ``} ` `    ``} ` ` `  `    ``// Driver program to test above functions ` `    ``public` `static` `void` `Main(``string``[] args) ` `    ``{ ` `        ``// Let us create binary tree given in the above ` `        ``// example ` `        ``Node root = ``new` `Node(1); ` `        ``root.left = ``new` `Node(2); ` `        ``root.right = ``new` `Node(3); ` `        ``root.left.left = ``new` `Node(4); ` `        ``root.left.right = ``new` `Node(5); ` `        ``root.right.left = ``new` `Node(6); ` `        ``root.right.right = ``new` `Node(7); ` `        ``root.left.left.left = ``new` `Node(8); ` `        ``root.right.left.left = ``new` `Node(9); ` `        ``root.right.left.right = ``new` `Node(10); ` ` `  `        ``int` `turn = 0; ` `        ``if` `((turn = NumberOfTurn(root, 5, 10)) != 0) ` `        ``{ ` `            ``Console.WriteLine(turn); ` `        ``} ` `        ``else` `        ``{ ` `            ``Console.WriteLine(``"Not Possible"``); ` `        ``} ` `    ``} ` ` `  `} ` ` `  `// This code is contributed by Shrikant13 `

Output:

```4
```

Time Complexity : O(n)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

This article is attributed to GeeksforGeeks.org

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