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.
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++
/* 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; } |
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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 |
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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 |
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C#
// A C# Program to find distance between // n1 and n2 using one traversal using System; class GFG { // (To the moderator) in c++ solution these // variables are declared as pointers hence //changes made to them reflects in the whole program // Global static variable public static int d1 = -1; public static int d2 = -1; public static int dist = 0; // 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; } } // Returns level of key k if it is present // in tree, otherwise returns -1 public 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 public 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. public 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 Code 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); Console.WriteLine("Dist(4, 5) = " + findDistance(root, 4, 5)); Console.WriteLine("Dist(4, 6) = " + findDistance(root, 4, 6)); Console.WriteLine("Dist(3, 4) = " + findDistance(root, 3, 4)); Console.WriteLine("Dist(2, 4) = " + findDistance(root, 2, 4)); Console.WriteLine("Dist(8, 5) = " + findDistance(root, 8, 5)); } } // This code is contributed by Shrikant13 |
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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.
C++
/* 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; } Node* LCA(Node * root, int n1,int n2) { // Your code here 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) { // Your code here 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; } |
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Java
/* Java Program to find distance between n1 and n2 using one traversal */public class GFG { public static class Node { int value; Node left; Node right; public Node(int value) { this.value = value; } } public static Node LCA(Node root, int n1, int n2) { if (root == null) return root; if (root.value == n1 || root.value == 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); else return LCA(root.right, n1, n2); } // Returns level of key k if it is present in // tree, otherwise returns -1 public static int findLevel(Node root, int a, int level) { if (root == null) return -1; if (root.value == a) return level; int left = findLevel(root.left, a, level + 1); if (left == -1) return findLevel(root.right, a, level + 1); return left; } public static 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 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 Srinivasan Jayaraman. |
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Python3
""" A python program to find distance between n1 and n2 in binary tree """# binary tree node class Node: # Constructor to create new node def __init__(self, data): self.data = data self.left = self.right = None # This function returns pointer to LCA of # two given values n1 and n2. def LCA(root, n1, n2): # Base case if root is None: return None # If either n1 or n2 matches with root's # key, report the presence by returning # root if root.data == n1 or root.data == n2: return root # Look for keys in left and right subtrees left = LCA(root.left, n1, n2) right = LCA(root.right, n1, n2) if left is not None and right is not None: return root # Otherwise check if left subtree or # right subtree is LCA if left: return left else: return right # function to find distance of any node # from root def findLevel(root, data, d, level): # Base case when tree is empty if root is None: return # Node is found then append level # value to list and return if root.data == data: d.append(level) return findLevel(root.left, data, d, level + 1) findLevel(root.right, data, d, level + 1) # function to find distance between two # nodes in a binary tree def findDistance(root, n1, n2): lca = LCA(root, n1, n2) # to store distance of n1 from lca d1 = [] # to store distance of n2 from lca d2 = [] # if lca exist if lca: # distance of n1 from lca findLevel(lca, n1, d1, 0) # distance of n2 from lca findLevel(lca, n2, d2, 0) return d1[0] + d2[0] else: return -1 # Driver program to test above function root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) root.right.left = Node(6) root.right.right = Node(7) root.right.left.right = Node(8) print("Dist(4,5) = ", findDistance(root, 4, 5)) print("Dist(4,6) = ", findDistance(root, 4, 6)) print("Dist(3,4) = ", findDistance(root, 3, 4)) print("Dist(2,4) = ", findDistance(root, 2, 4)) print("Dist(8,5) = ", findDistance(root, 8, 5)) # This article is contributed by Shweta Singh. |
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C#
using System; /* C# Program to find distance between n1 and n2 using one traversal */public class GFG { public class Node { public int value; public Node left; public Node right; public Node(int value) { this.value = value; } } public static Node LCA(Node root, int n1, int n2) { if (root == null) { return root; } if (root.value == n1 || root.value == 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); } else { return LCA(root.right, n1, n2); } } // Returns level of key k if it is present in // tree, otherwise returns -1 public static int findLevel(Node root, int a, int level) { if (root == null) { return -1; } if (root.value == a) { return level; } int left = findLevel(root.left, a, level + 1); if (left == -1) { return findLevel(root.right, a, level + 1); } return left; } public static 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 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); Console.WriteLine("Dist(4, 5) = " + findDistance(root, 4, 5)); Console.WriteLine("Dist(4, 6) = " + findDistance(root, 4, 6)); Console.WriteLine("Dist(3, 4) = " + findDistance(root, 3, 4)); Console.WriteLine("Dist(2, 4) = " + findDistance(root, 2, 4)); Console.WriteLine("Dist(8, 5) = " + findDistance(root, 8, 5)); } } // This code is contributed by Shrikant13 |
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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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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