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.
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 <bits/stdc++.h>using namespace std;// node classstruct 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 Treestruct 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 treesNode* 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 Codeint 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 |
Python3
# Python 3 program for disconnecting a# node to result in three different treeclass 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 @staticmethod def findLCA( root, n1, n2) : # Base case if (root == None) : 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 or root.key == n2) : return root # Look for keys in left and right subtrees left_lca = RemoveEdge.findLCA(root.left, n1, n2) right_lca = RemoveEdge.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 != None and right_lca != None) : return root # Otherwise check if left subtree or right subtree is LCA return left_lca if (left_lca != None) else 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 @staticmethod def findNode( root, a, b, c) : # lca of a, b x = RemoveEdge.findLCA(root, a, b) # lca of b, c y = RemoveEdge.findLCA(root, b, c) # lca of c, a z = RemoveEdge.findLCA(root, c, a) if (x.key == y.key) : return z elif(x.key == z.key) : return y else : return x @staticmethod def main( args) : root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) root.left.left.left = Node(8) root.left.left.right = Node(9) root.left.right.left = Node(10) root.left.right.right = Node(11) root.right.left = Node(6) root.right.right = Node(7) root.right.left.left = Node(12) root.right.left.right = Node(13) root.right.right.left = Node(14) root.right.right.right = Node(15) print("Disconnect node " + str(RemoveEdge.findNode(root, 5, 6, 15).key) + " from the tree", end ="")# Node classclass Node : key = 0 left = None right = None def __init__(self, data) : self.key = data if __name__=="__main__": RemoveEdge.main([]) # This code is contributed by aadityaburujwale. |
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 |
Javascript
// node classclass Node { constructor(key) { this.key = key; this.left = null; this.right = null; }}function newNode(key) { let temp = new Node(key); return temp;}// LCA functionfunction findLCA(root, n1, 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 let left_lca = findLCA(root.left, n1, n2); let 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 treesfunction findNode(root, a, b, c) { // lca of a, b let x = findLCA(root, a, b); // lca of b, c let y = findLCA(root, b, c); // lca of c, a let z = findLCA(root, c, a); if (x.key == y.key) { return z; } else if (x.key == z.key) { return y; } else { return x; }}// Declare tree// Insert elements in the treelet 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 ) console.log("Disconnect node" , findNode(root, 5, 6, 15).key , "from the tree");// This code is contributed by akashish__ |
Output:
Disconnect node 3 from the tree
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