Remove edges connected to a node such that the three given nodes are in different trees

Given a binary tree and 3 nodes a, b and c, the task is to find a node in the tree such that after removing all the edge connected to that node, a, b and c are in three different trees.

Given below is a tree with input nodes as c, j and o. In the above tree, if node i gets disconnected from the tree, then the given nodes c, j, and o will be in three different trees which have been shown below. Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple approach is to find LCA of all possible pairs of nodes given.
Let,

• lca of ( a, b) = x
• lca of (b, c) = y
• lca of (c, a) = z

In any case, either of (x, y), (y, z), (z, x) or (x, y, z) will always be the same. In the first three cases, return the node which is not the same. In the last case returning any node of x, y or z will give the answer.

Below is the implementation of the above approach:

C++

 // C++ program for disconnecting a // node to result in three different tree #include using namespace std;    // node class struct Node {     int key;     struct Node *left, *right; }; Node* newNode(int key) {     Node* temp = new Node;     temp->key = key;     temp->left = temp->right = NULL;     return (temp); }    // LCA function taken from the above link mentioned // This function returns a pointer to LCA of two given // values n1 and n2. This function assumes that n1 and n2 // are present in Binary Tree 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; }    // the function assumes a, b, c are present in the tree // and returns a node disconnecting which // results in all three nodes in different trees Node* findNode(Node* root, int a, int b, int c) {     // lca of a, b     Node* x = findLCA(root, a, b);        // lca of b, c     Node* y = findLCA(root, b, c);        // lca of c, a     Node* z = findLCA(root, c, a);        if (x->key == y->key)         return z;     else if (x->key == z->key)         return y;     else         return x; }    // Driver Code int main() {     // Declare tree     // Insert elements in the tree     Node* root = newNode(1);        root->left = newNode(2);     root->right = newNode(3);        root->left->left = newNode(4);     root->left->right = newNode(5);        root->left->left->left = newNode(8);     root->left->left->right = newNode(9);        root->left->right->left = newNode(10);     root->left->right->right = newNode(11);        root->right->left = newNode(6);     root->right->right = newNode(7);     root->right->left->left = newNode(12);     root->right->left->right = newNode(13);     root->right->right->left = newNode(14);     root->right->right->right = newNode(15);        /*             1         /     \        2       3      /  \     /  \     4   5     6    7    /\  / \   / \  / \   8 9 10 11 12 13 14 15                                 */        // update all the suitable_children     // keys of all the nodes in O( N )        cout << "Disconnect node "          << findNode(root, 5, 6, 15)->key          << " from the tree";        return 0; }

Java

 // Java program for disconnecting a  // node to result in three different tree  public class RemoveEdge {        // LCA function taken from the above link mentioned      // This function returns a pointer to LCA of two given      // values n1 and n2. This function assumes that n1 and n2      // are present in Binary Tree      static Node findLCA(Node root, int n1, int n2)      {          // Base case          if (root == null)              return root;               // 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;      }           // the function assumes a, b, c are present in the tree      // and returns a node disconnecting which      // results in all three nodes in different trees      static Node findNode(Node root, int a, int b, int c)      {          // lca of a, b          Node x = findLCA(root, a, b);          // lca of b, c          Node y = findLCA(root, b, c);          // lca of c, a          Node z = findLCA(root, c, a);               if (x.key == y.key)              return z;          else if (x.key == z.key)              return y;          else             return x;      }         public static void main(String args[]) {         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.left.left.left = new Node(8);          root.left.left.right = new Node(9);          root.left.right.left = new Node(10);          root.left.right.right = new Node(11);          root.right.left = new Node(6);          root.right.right = new Node(7);          root.right.left.left = new Node(12);          root.right.left.right = new Node(13);          root.right.right.left = new Node(14);          root.right.right.right = new Node(15);          System.out.print("Disconnect node "+findNode(root, 5, 6, 15).key+" from the tree");     } }    // Node class  class Node {      int key;      Node left, right;      Node (int data)     {         this.key=data;     } };  //This code is contributed by Gaurav Tiwari

C#

 // C# program for disconnecting a  // node to result in three different tree  using System;    public class RemoveEdge {         // LCA function taken from the      // above link mentioned This function     // returns a pointer to LCA of two given      // values n1 and n2. This function     // assumes that n1 and n2      // are present in Binary Tree      static Node findLCA(Node root, int n1, int n2)      {          // Base case          if (root == null)              return root;                 // 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;      }             // the function assumes a, b, c      // are present in the tree and returns     // a node disconnecting which results     // in all three nodes in different trees      static Node findNode(Node root, int a, int b, int c)      {          // lca of a, b          Node x = findLCA(root, a, b);                     // lca of b, c          Node y = findLCA(root, b, c);                     // lca of c, a          Node z = findLCA(root, c, a);                 if (x.key == y.key)              return z;          else if (x.key == z.key)              return y;          else             return x;      }         // Driver code     public static void Main(String []args)      {          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.left.left.left = new Node(8);          root.left.left.right = new Node(9);          root.left.right.left = new Node(10);          root.left.right.right = new Node(11);          root.right.left = new Node(6);          root.right.right = new Node(7);          root.right.left.left = new Node(12);          root.right.left.right = new Node(13);          root.right.right.left = new Node(14);          root.right.right.right = new Node(15);          Console.Write("Disconnect node "+                         findNode(root, 5, 6, 15).key+                         " from the tree");      }  }     // Node class  public class Node  {      public int key;      public Node left, right;      public Node (int data)      {          this.key=data;      }  };     // This code contributed by Rajput-Ji

Output:

Disconnect node 3 from the tree

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Improved By : _Gaurav_Tiwari, Rajput-Ji

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