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Print common nodes on path from root (or common ancestors)

Given a binary tree and two nodes, the task is to Print all the nodes that are common for 2 given nodes in a binary tree.

Examples: 

Given binary tree is :
                     1
                  /    \
                2       3
              /   \     /  \
             4     5   6    7
            /        /  \
           8        9   10

Given nodes 9 and 7, so the common nodes are:-
1, 3

Asked in : Amazon

  1. Find the LCA of given two nodes.
  2. Print all ancestors of the LCA as done in this post, also print the LCA.

Implementation:




// C++ Program to find common nodes for given two nodes
#include <bits/stdc++.h>
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;
}
  
// Utility Function to print all ancestors of LCA
bool printAncestors(struct Node* root, int target)
{
    /* base cases */
    if (root == NULL)
        return false;
  
    if (root->key == target) {
        cout << root->key << " ";
        return true;
    }
  
    /* If target is present in either left or right
      subtree of this node, then print this node */
    if (printAncestors(root->left, target) ||
        printAncestors(root->right, target)) {
        cout << root->key << " ";
        return true;
    }
  
    /* Else return false */
    return false;
}
  
// Function to find nodes common to given two nodes
bool findCommonNodes(struct Node* root, int first,
                                       int second)
{
    struct Node* LCA = findLCA(root, first, second);
    if (LCA == NULL)
        return false;
  
    printAncestors(root, LCA->key);
}
  
// 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);
  
    if (findCommonNodes(root, 9, 7) == false)
        cout << "No Common nodes";
  
    return 0;
}




// Java Program to find common nodes for given two nodes
import java.util.LinkedList;
   
// Class to represent Tree node 
class Node 
{
    int data;
    Node left, right;
   
    public Node(int item) 
    {
        data = item;
        left = null;
        right = null;
    }
}
   
// Class to count full nodes of Tree 
class BinaryTree 
{
    static Node root;
// 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.data == n1 || root.data == 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;
}
   
// Utility Function to print all ancestors of LCA
static boolean printAncestors(Node root, int target)
{
    /* base cases */
    if (root == null)
        return false;
   
    if (root.data == target) {
        System.out.print(root.data+ " ");
        return true;
    }
   
    /* If target is present in either left or right
    subtree of this node, then print this node */
    if (printAncestors(root.left, target) ||
        printAncestors(root.right, target)) {
        System.out.print(root.data+ " ");
        return true;
    }
   
    /* Else return false */
    return false;
}
   
// Function to find nodes common to given two nodes
static boolean findCommonNodes(Node root, int first,
                                    int second)
{
    Node LCA = findLCA(root, first, second);
    if (LCA == null)
        return false;
   
    printAncestors(root, LCA.data);
    return true;
}
   
// Driver program to test above functions
    public static void main(String args[]) 
    {
    /*Let us create Binary Tree shown in 
        above example */
   
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);
        tree.root.right.left = new Node(6);
        tree.root.right.right = new Node(7);
        tree.root.left.left.left = new Node(8);
        tree.root.right.left.left = new Node(9);
        tree.root.right.left.right = new Node(10);
   
   if (findCommonNodes(root, 9, 7) == false)
    System.out.println("No Common nodes");
   
    }
}
  
// This code is contributed by Mr Somesh Awasthi




# Python3 Program to find common
# nodes for given two nodes 
  
# Utility class to create a new tree Node 
class newNode:
    def __init__(self, key):
        self.key = key 
        self.left = self.right = None
      
# Utility function to find the LCA of
# two given values n1 and n2. 
def findLCA(root, n1, n2):
      
    # Base case 
    if (root == None):
        return None
  
    # 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 = findLCA(root.left, n1, n2) 
    right_lca = findLCA(root.right, n1, n2) 
  
    # If both of the above calls return Non-None, 
    # then one key is present in once subtree and 
    # other is present in other, So this node is the LCA 
    if (left_lca and right_lca): 
        return root 
  
    # Otherwise check if left subtree or 
    # right subtree is LCA
    if (left_lca != None):
        return left_lca
    else:
        return right_lca 
  
# Utility Function to print all ancestors of LCA 
def printAncestors(root, target):
      
    # base cases 
    if (root == None):
        return False
  
    if (root.key == target): 
        print(root.key, end = " "
        return True
  
    # If target is present in either left or right 
    # subtree of this node, then print this node 
    if (printAncestors(root.left, target) or 
        printAncestors(root.right, target)):
            print(root.key, end = " "
            return True
  
    # Else return False 
    return False
  
# Function to find nodes common to given two nodes 
def findCommonNodes(root, first, second):
    LCA = findLCA(root, first, second) 
    if (LCA == None):
        return False
  
    printAncestors(root, LCA.key)
  
# Driver Code
if __name__ == '__main__':
  
    # Let us create binary tree given 
    # in the above example 
    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
  
    if (findCommonNodes(root, 9, 7) == False): 
        print("No Common nodes")
  
# This code is contributed by PranchalK




using System;
  
// C# Program to find common nodes for given two nodes 
  
// Class to represent Tree node 
public class Node
{
    public int data;
    public Node left, right;
  
    public Node(int item)
    {
        data = item;
        left = null;
        right = null;
    }
}
  
// Class to count full nodes of Tree 
public class BinaryTree
{
    public static Node root;
// 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.data == n1 || root.data == 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;
}
  
// Utility Function to print all ancestors of LCA 
public static bool printAncestors(Node root, int target)
{
    /* base cases */
    if (root == null)
    {
        return false;
    }
  
    if (root.data == target)
    {
        Console.Write(root.data + " ");
        return true;
    }
  
    /* If target is present in either left or right 
    subtree of this node, then print this node */
    if (printAncestors(root.left, target) 
    || printAncestors(root.right, target))
    {
        Console.Write(root.data + " ");
        return true;
    }
  
    /* Else return false */
    return false;
}
  
// Function to find nodes common to given two nodes 
public static bool findCommonNodes(Node root, 
                            int first, int second)
{
    Node LCA = findLCA(root, first, second);
    if (LCA == null)
    {
        return false;
    }
  
    printAncestors(root, LCA.data);
    return true;
}
  
// Driver program to test above functions 
    public static void Main(string[] args)
    {
    /*Let us create Binary Tree shown in 
        above example */
  
        BinaryTree tree = new BinaryTree();
        BinaryTree.root = new Node(1);
        BinaryTree.root.left = new Node(2);
        BinaryTree.root.right = new Node(3);
        BinaryTree.root.left.left = new Node(4);
        BinaryTree.root.left.right = new Node(5);
        BinaryTree.root.right.left = new Node(6);
        BinaryTree.root.right.right = new Node(7);
        BinaryTree.root.left.left.left = new Node(8);
        BinaryTree.root.right.left.left = new Node(9);
        BinaryTree.root.right.left.right = new Node(10);
  
if (findCommonNodes(root, 9, 7) == false)
{
    Console.WriteLine("No Common nodes");
}
  
    }
}
  
// This code is contributed by Shrikant13




<script>
  
// Javascript program to find common 
// nodes for given two nodes
class Node
{
    constructor(data)
    {
        this.left = null;
        this.right = null;
        this.data = data;
    }
}
  
let root;
  
// Utility function to find the LCA of
// two given values n1 and n2.
function 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.data == n1 || root.data == 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!=null && right_lca!=null)
        return root;
  
    // Otherwise check if left subtree or right
    // subtree is LCA
    return (left_lca != null) ? left_lca : right_lca;
}
  
// Utility Function to print all ancestors of LCA
function printAncestors(root, target)
{
      
    // Base cases 
    if (root == null)
        return false;
  
    if (root.data == target) 
    {
        document.write(root.data + " ");
        return true;
    }
  
    // If target is present in either left or right
    // subtree of this node, then print this node 
    if (printAncestors(root.left, target) ||
        printAncestors(root.right, target))
    {
        document.write(root.data+ " ");
        return true;
    }
  
    // Else return false 
    return false;
}
  
// Function to find nodes common to given two nodes
function findCommonNodes(root, first, second)
{
    let LCA = findLCA(root, first, second);
    if (LCA == null)
        return false;
  
    printAncestors(root, LCA.data);
    return true;
}
  
// Driver code
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);
  
if (findCommonNodes(root, 9, 7) == false)
    document.write("No Common nodes");
      
// This code is contributed by divyeshrabadiya07
  
</script>

Output
3 1 

Time Complexity: O(N) where N is the number of nodes in given Binary Tree

Auxiliary Space: O(h) where h is the height of binary tree due to recursion stack call.

 


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