Find distance between two nodes of a Binary Tree

Find the distance between two keys in a binary tree, no parent pointers are given. Distance between two nodes is the minimum number of edges to be traversed to reach one node from other.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The distance between two nodes can be obtained in terms of lowest common ancestor. Following is the formula.

```Dist(n1, n2) = Dist(root, n1) + Dist(root, n2) - 2*Dist(root, lca)
'n1' and 'n2' are the two given keys
'root' is root of given Binary Tree.
'lca' is lowest common ancestor of n1 and n2
Dist(n1, n2) is the distance between n1 and n2.
```

Following is the implementation of above approach. The implementation is adopted from last code provided in Lowest Common Ancestor Post.

C++

```/* Program to find distance between n1 and n2 using
one traversal */
#include <iostream>
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;
}

// Returns level of key k if it is present in tree,
// otherwise returns -1
int findLevel(Node *root, int k, int level)
{
// Base Case
if (root == NULL)
return -1;

// If key is present at root, or in left subtree
// or right subtree, return true;
if (root->key == k)
return level;

int l = findLevel(root->left, k, level+1);
return (l != -1)? l : findLevel(root->right, k, level+1);
}

// This function returns pointer to LCA of two given
// values n1 and n2. It also sets d1, d2 and dist if
// one key is not ancestor of other
// d1 --> To store distance of n1 from root
// d2 --> To store distance of n2 from root
// lvl --> Level (or distance from root) of current node
// dist --> To store distance between n1 and n2
Node *findDistUtil(Node* root, int n1, int n2, int &d1,
int &d2, int &dist, int lvl)
{
// 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)
{
d1 = lvl;
return root;
}
if (root->key == n2)
{
d2 = lvl;
return root;
}

// Look for n1 and n2 in left and right subtrees
Node *left_lca  = findDistUtil(root->left, n1, n2,
d1, d2, dist, lvl+1);
Node *right_lca = findDistUtil(root->right, n1, n2,
d1, d2, dist, lvl+1);

// 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)
{
dist = d1 + d2 - 2*lvl;
return root;
}

// Otherwise check if left subtree or right subtree
// is LCA
return (left_lca != NULL)? left_lca: right_lca;
}

// The main function that returns distance between n1
// and n2. This function returns -1 if either n1 or n2
// is not present in Binary Tree.
int findDistance(Node *root, int n1, int n2)
{
// Initialize d1 (distance of n1 from root), d2
// (distance of n2 from root) and dist(distance
// between n1 and n2)
int d1 = -1, d2 = -1, dist;
Node *lca = findDistUtil(root, n1, n2, d1, d2,
dist, 1);

// If both n1 and n2 were present in Binary
// Tree, return dist
if (d1 != -1 && d2 != -1)
return dist;

// If n1 is ancestor of n2, consider n1 as root
// and find level of n2 in subtree rooted with n1
if (d1 != -1)
{
dist = findLevel(lca, n2, 0);
return dist;
}

// If n2 is ancestor of n1, consider n2 as root
// and find level of n1 in subtree rooted with n2
if (d2 != -1)
{
dist = findLevel(lca, n1, 0);
return dist;
}

return -1;
}

// 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->right->left->right = newNode(8);
cout << "Dist(4, 5) = " << findDistance(root, 4, 5);
cout << "nDist(4, 6) = " << findDistance(root, 4, 6);
cout << "nDist(3, 4) = " << findDistance(root, 3, 4);
cout << "nDist(2, 4) = " << findDistance(root, 2, 4);
cout << "nDist(8, 5) = " << findDistance(root, 8, 5);
return 0;
} ```

Java

```/* A Java Program to find distance between n1 and n2
using one traversal */
public class DistanceBetweenTwoKey
{
// (To the moderator) in c++ solution this variable
// are declared as pointers hence changes made to them
// reflects in the whole program

// Global static variable
static int d1 = -1;
static int d2 = -1;
static int dist = 0;

// A Binary Tree Node
static class Node{
Node left, right;
int key;

// constructor
Node(int key){
this.key = key;
left = null;
right = null;
}
}

// Returns level of key k if it is present in tree,
// otherwise returns -1
static int findLevel(Node root, int k, int level)
{
// Base Case
if (root == null)
return -1;

// If key is present at root, or in left subtree or right subtree,
// return true;
if (root.key == k)
return level;

int l = findLevel(root.left, k, level + 1);
return (l != -1)? l : findLevel(root.right, k, level + 1);
}

// This function returns pointer to LCA of two given values n1 and n2.
// It also sets d1, d2 and dist if one key is not ancestor of other
// d1 --> To store distance of n1 from root
// d2 --> To store distance of n2 from root
// lvl --> Level (or distance from root) of current node
// dist --> To store distance between n1 and n2
static Node findDistUtil(Node root, int n1, int n2, int lvl){

// 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){
d1 = lvl;
return root;
}
if (root.key == n2)
{
d2 = lvl;
return root;
}

// Look for n1 and n2 in left and right subtrees
Node left_lca = findDistUtil(root.left, n1, n2,  lvl + 1);
Node right_lca = findDistUtil(root.right, n1, n2,  lvl + 1);

// 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)
{
dist = (d1 + d2) - 2*lvl;
return root;
}

// Otherwise check if left subtree or right subtree is LCA
return (left_lca != null)? left_lca : right_lca;
}

// The main function that returns distance between n1 and n2
// This function returns -1 if either n1 or n2 is not present in
// Binary Tree.
static int findDistance(Node root, int n1, int n2){
d1 = -1;
d2 = -1;
dist = 0;
Node lca = findDistUtil(root, n1, n2, 1);

// If both n1 and n2 were present in Binary Tree, return dist
if (d1 != -1 && d2 != -1)
return dist;

// If n1 is ancestor of n2, consider n1 as root and find level
// of n2 in subtree rooted with n1
if (d1 != -1)
{
dist = findLevel(lca, n2, 0);
return dist;
}

// If n2 is ancestor of n1, consider n2 as root and find level
// of n1 in subtree rooted with n2
if (d2 != -1)
{
dist = findLevel(lca, n1, 0);
return dist;
}

return -1;
}

// 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.right.left.right = new Node(8);

System.out.println("Dist(4, 5) = "+findDistance(root, 4, 5));
System.out.println("Dist(4, 6) = "+findDistance(root, 4, 6));
System.out.println("Dist(3, 4) = "+findDistance(root, 3, 4));
System.out.println("Dist(2, 4) = "+findDistance(root, 2, 4));
System.out.println("Dist(8, 5) = " +findDistance(root, 8, 5));

}
}
// This code is contributed by Sumit Ghosh
```

Python

```# Python Program to find distance between
# n1 and n2 using one traversal

class Node:
def __init__(self, data):
self.data = data
self.right = None
self.left = None

def pathToNode(root, path, k):

# base case handling
if root is None:
return False

# append the node value in path
path.append(root.data)

# See if the k is same as root's data
if root.data == k :
return True

# Check if k is found in left or right
# sub-tree
if ((root.left != None and pathToNode(root.left, path, k)) or
(root.right!= None and pathToNode(root.right, path, k))):
return True

# If not present in subtree rooted with root,
# remove root from path and return False
path.pop()
return False

def distance(root, data1, data2):
if root:
# store path corresponding to node: data1
path1 = []
pathToNode(root, path1, data1)

# store path corresponding to node: data2
path2 = []
pathToNode(root, path2, data2)

# iterate through the paths to find the
# common path length
i=0
while i<len(path1) and i<len(path2):
# get out as soon as the path differs
# or any path's length get exhausted
if path1[i] != path2[i]:
break
i = i+1

# get the path length by deducting the
# intersecting path length (or till LCA)
return (len(path1)+len(path2)-2*i)
else:
return 0

# Driver Code to test above functions
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.right.right= Node(7)
root.right.left = Node(6)
root.left.right = Node(5)
root.right.left.right = Node(8)

dist = distance(root, 4, 5)
print "Distance between node {} & {}: {}".format(4, 5, dist)

dist = distance(root, 4, 6)
print "Distance between node {} & {}: {}".format(4, 6, dist)

dist = distance(root, 3, 4)
print "Distance between node {} & {}: {}".format(3, 4, dist)

dist = distance(root, 2, 4)
print "Distance between node {} & {}: {}".format(2, 4, dist)

dist = distance(root, 8, 5)
print "Distance between node {} & {}: {}".format(8, 5, dist)

# This program is contributed by Aartee
```

Output:
```Dist(4, 5) = 2
Dist(4, 6) = 4
Dist(3, 4) = 3
Dist(2, 4) = 1
Dist(8, 5) = 5```

Time Complexity: Time complexity of the above solution is O(n) as the method does a single tree traversal.

Thanks to Atul Singh for providing the initial solution for this post.

Better Solution :
We first find LCA of two nodes. Then we find distance from LCA to two nodes.

```/* Program to find distance between n1 and n2
using one traversal */
#include <iostream>
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;
}
Node* LCA(Node * root, int n1,int n2)
{
if (root == NULL)
return root;
if (root->key == n1 || root->key == n2)
return root;

Node* left = LCA(root->left, n1, n2);
Node* right = LCA(root->right, n1, n2);

if (left != NULL && right != NULL)
return root;
if (left != NULL)
return LCA(root->left, n1, n2);

return LCA(root->right, n1, n2);
}

// Returns level of key k if it is present in
// tree, otherwise returns -1
int findLevel(Node *root, int k, int level)
{
if(root == NULL) return -1;
if(root->key == k) return level;

int left = findLevel(root->left, k, level+1);
if (left == -1)
return findLevel(root->right, k, level+1);
return left;
}

int findDistance(Node* root, int a, int b)
{
Node* lca = LCA(root, a , b);

int d1 = findLevel(lca, a, 0);
int d2 = findLevel(lca, b, 0);

return d1 + d2;
}

// 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->right->left->right = newNode(8);
cout << "Dist(4, 5) = " << findDistance(root, 4, 5);
cout << "\nDist(4, 6) = " << findDistance(root, 4, 6);
cout << "\nDist(3, 4) = " << findDistance(root, 3, 4);
cout << "\nDist(2, 4) = " << findDistance(root, 2, 4);
cout << "\nDist(8, 5) = " << findDistance(root, 8, 5);
return 0;
}
```

Output :

```Dist(4, 5) = 2
Dist(4, 6) = 4
Dist(3, 4) = 3
Dist(2, 4) = 1
Dist(8, 5) = 5```

Thanks to NILMADHAB MONDAL for suggesting this solution.

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