Print the path common to the two paths from the root to the two given nodes

Given a binary tree with distinct nodes(no two nodes have the same have data values). The problem is to print the path common to the two paths from the root to the two given nodes n1 and n2. If either of the nodes are not present then print “No Common Path”.

Examples:

Input :          1
               /   \
              2     3
             / \   /  \
            4   5  6   7
               /    \   
              8      9

          n1 = 4, n2 = 8

Output : 1->2
Path form root to n1:
1->2->4

Path form root to n2:
1->2->5->8

Common Path:
1->2

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:The following steps are:

Find the LCA(Lowest Common Ancestor) of the two nodes n1 and n2. Refer this.
If LCA exits then print the path from the root to LCA. Refer this. Else print “No Common Path”.

C++

// C++ implementation to print the path common to the  
// two paths from the root to the two given nodes 
#include <bits/stdc++.h> 
  
using namespace std; 
  
// structure of a node of binary tree 
struct Node 
{ 
    int data; 
    Node *left, *right; 
}; 
  
/* Helper function that allocates a new node with the 
   given data and NULL left and right pointers. */
struct Node* getNode(int data) 
{ 
    struct Node *newNode = (struct Node*)malloc(sizeof(struct Node)); 
    newNode->data = data; 
    newNode->left = newNode->right = NULL; 
    return newNode; 
} 
  
// This function returns pointer to LCA of two given values n1 and n2. 
// v1 is set as true by this function if n1 is found 
// v2 is set as true by this function if n2 is found 
struct Node *findLCAUtil(struct Node* root, int n1, int n2, bool &v1, bool &v2) 
{ 
    // Base case 
    if (root == NULL) return NULL; 
   
    // If either n1 or n2 matches with root's data, report the presence 
    // by setting v1 or v2 as true and return root (Note that if a key 
    // is ancestor of other, then the ancestor key becomes LCA) 
    if (root->data == n1) 
    { 
        v1 = true; 
        return root; 
    } 
    if (root->data == n2) 
    { 
        v2 = true; 
        return root; 
    } 
   
    // Look for nodes in left and right subtrees 
    Node *left_lca  = findLCAUtil(root->left, n1, n2, v1, v2); 
    Node *right_lca = findLCAUtil(root->right, n1, n2, v1, v2); 
   
    // If both of the above calls return Non-NULL, then one node 
    // is present in one subtree and other is present in other, 
    // So this current 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; 
} 
  
// Returns true if key k is present in tree rooted with root 
bool find(Node *root, int k) 
{ 
    // Base Case 
    if (root == NULL) 
        return false; 
   
    // If key k is present at root, or in left subtree  
    // or right subtree, return true 
    if (root->data == k || find(root->left, k) ||  find(root->right, k)) 
        return true; 
   
    // Else return false 
    return false; 
} 
  
// This function returns LCA of n1 and n2 only if both n1 and n2  
// are present in tree, otherwise returns NULL 
Node *findLCA(Node *root, int n1, int n2) 
{ 
    // Initialize n1 and n2 as not visited 
    bool v1 = false, v2 = false; 
   
    // Find lca of n1 and n2 
    Node *lca = findLCAUtil(root, n1, n2, v1, v2); 
   
    // Return LCA only if both n1 and n2 are present in tree 
    if (v1 && v2 || v1 && find(lca, n2) || v2 && find(lca, n1)) 
        return lca; 
   
    // Else return NULL 
    return NULL; 
} 
  
// function returns true if  
// there is a path from root to  
// the given node. It also populates  
// 'arr' with the given path 
bool hasPath(Node *root, vector<int>& arr, int x) 
{ 
    // if root is NULL 
    // there is no path 
    if (!root) 
        return false; 
      
    // push the node's value in 'arr' 
    arr.push_back(root->data);     
      
    // if it is the required node 
    // return true 
    if (root->data == x)     
        return true; 
      
    // else check whether there    the required node lies in the 
    // left subtree or right subtree of the current node 
    if (hasPath(root->left, arr, x) || 
        hasPath(root->right, arr, x)) 
        return true; 
      
    // required node does not lie either in the  
    // left or right subtree of the current node 
    // Thus, remove current node's value from 'arr' 
    // and then return false;     
    arr.pop_back(); 
    return false;             
} 
  
// function to print the path common 
// to the two paths from the root  
// to the two given nodes if the nodes  
// lie in the binary tree 
void printCommonPath(Node *root, int n1, int n2) 
{ 
    // vector to store the common path 
    vector<int> arr; 
      
    // LCA of node n1 and n2 
    Node *lca = findLCA(root, n1, n2); 
      
    // if LCA of both n1 and n2 exists 
    if (lca) 
    { 
        // then print the path from root to 
        // LCA node 
        if (hasPath(root, arr, lca->data)) 
        { 
            for (int i=0; i<arr.size()-1; i++)     
                cout << arr[i] << "->"; 
            cout << arr[arr.size() - 1];     
        }     
    } 
      
    // LCA is not present in the binary tree  
    // either n1 or n2 or both are not present 
    else
        cout << "No Common Path"; 
} 
  
// Driver program to test above 
int main() 
{ 
    // binary tree formation 
    struct Node *root = getNode(1); 
    root->left = getNode(2); 
    root->right = getNode(3); 
    root->left->left = getNode(4); 
    root->left->right = getNode(5); 
    root->right->left = getNode(6); 
    root->right->right = getNode(7); 
    root->left->right->left = getNode(8); 
    root->right->left->right = getNode(9); 
          
    int n1 = 4, n2 = 8; 
    printCommonPath(root, n1, n2); 
    return 0; 
} 

Java

// Java implementation to print the path common to the   
// two paths from the root to the two given nodes  
import java.util.ArrayList; 
public class PrintCommonPath { 
  
    // Initialize n1 and n2 as not visited  
    static boolean v1 = false, v2 = false;  
  
    // This function returns pointer to LCA of two given  
    // values n1 and n2. This function assumes that n1 and  
    // n2 are present in Binary Tree  
    static Node findLCAUtil(Node node, int n1, int n2)  
    {  
        // Base case  
        if (node == null)  
            return null;  
            
        //Store result in temp, in case of key match so that we can search for other key also.  
        Node temp=null;  
    
        // If either n1 or n2 matches with root's key, report the presence  
        // by setting v1 or v2 as true and return root (Note that if a key  
        // is ancestor of other, then the ancestor key becomes LCA)  
        if (node.data == n1)  
        {  
            v1 = true;  
            temp = node;  
        }  
        if (node.data == n2)  
        {  
            v2 = true;  
            temp = node;  
        }  
    
        // Look for keys in left and right subtrees  
        Node left_lca = findLCAUtil(node.left, n1, n2);  
        Node right_lca = findLCAUtil(node.right, n1, n2);  
    
        if (temp != null)  
            return temp;  
    
        // 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 node;  
    
        // Otherwise check if left subtree or right subtree is LCA  
        return (left_lca != null) ? left_lca : right_lca;  
    } 
  
    // Returns true if key k is present in tree rooted with root  
    static boolean find(Node root, int k)  
    {  
        // Base Case  
        if (root == null)  
            return false;  
     
        // If key k is present at root, or in left subtree   
        // or right subtree, return true  
        if (root.data == k || find(root.left, k) ||  find(root.right, k))  
            return true;  
     
        // Else return false  
        return false;  
    }  
  
    // This function returns LCA of n1 and n2 only if both n1 and n2   
    // are present in tree, otherwise returns null  
    static Node findLCA(Node root, int n1, int n2)  
    {  
        // Find lca of n1 and n2  
        Node lca = findLCAUtil(root, n1, n2);  
     
        // Return LCA only if both n1 and n2 are present in tree  
        if (v1 && v2 || v1 && find(lca, n2) || v2 && find(lca, n1))  
            return lca;  
     
        // Else return null  
        return null;  
    }  
  
    // function returns true if   
    // there is a path from root to   
    // the given node. It also populates   
    // 'arr' with the given path  
    static boolean hasPath(Node root, ArrayList<Integer> arr, int x)  
    {  
        // if root is null  
        // there is no path  
        if (root==null)  
            return false;  
        
        // push the node's value in 'arr'  
        arr.add(root.data);      
        
        // if it is the required node  
        // return true  
        if (root.data == x)      
            return true;  
        
        // else check whether there    the required node lies in the  
        // left subtree or right subtree of the current node  
        if (hasPath(root.left, arr, x) ||  
            hasPath(root.right, arr, x))  
            return true;  
        
        // required node does not lie either in the   
        // left or right subtree of the current node  
        // Thus, remove current node's value from 'arr'  
        // and then return false;      
        arr.remove(arr.size()-1);  
        return false;              
    }  
  
    // function to print the path common  
    // to the two paths from the root   
    // to the two given nodes if the nodes   
    // lie in the binary tree  
    static void printCommonPath(Node root, int n1, int n2)  
    {  
        // ArrayList to store the common path  
        ArrayList<Integer> arr=new ArrayList<>(); 
        
        // LCA of node n1 and n2  
        Node lca = findLCA(root, n1, n2);  
        
        // if LCA of both n1 and n2 exists  
        if (lca!=null)  
        {    
            // then print the path from root to  
            // LCA node  
            if (hasPath(root, arr, lca.data))  
            {  
                for (int i=0; i<arr.size()-1; i++)      
                    System.out.print(arr.get(i)+"->"); 
                    System.out.print(arr.get(arr.size() - 1));      
            }      
        }  
        
        // LCA is not present in the binary tree   
        // either n1 or n2 or both are not present  
        else
            System.out.print("No Common Path"); 
    }  
  
    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.right.left = new Node(6);  
        root.right.right = new Node(7);  
        root.left.right.left = new Node(8);  
        root.right.left.right = new Node(9);  
            
        int n1 = 4, n2 = 8;  
        printCommonPath(root, n1, n2);  
        } 
} 
  
/* Class containing left and right child of current  
 node and key value*/
class Node  
{  
    int data;  
    Node left, right;  
    
    public Node(int item)  
    {  
        data = item;  
        left = right = null;  
    }  
}  
//This code is contributed by Gaurav Tiwari 

Python3

# Python implementation to print the path common to the 
# two paths from the root to the two given nodes 
  
# structure of a node of binary tree  
class Node: 
    def __init__(self, data): 
        self.data = data 
        self.left = None
        self.right = None
  
# This function returns pointer to LCA of two given values n1 and n2. 
# v1 is set as True by this function if n1 is found 
# v2 is set as True by this function if n2 is found 
def findLCAUtil(root: Node, n1: int, n2: int) -> Node: 
    global v1, v2 
  
    # Base case 
    if (root is None): 
        return None
  
    # If either n1 or n2 matches with root's data, report the presence 
    # by setting v1 or v2 as True and return root (Note that if a key 
    # is ancestor of other, then the ancestor key becomes LCA) 
    if (root.data == n1): 
        v1 = True
        return root 
  
    if (root.data == n2): 
  
        v2 = True
        return root 
  
    # Look for nodes in left and right subtrees 
    left_lca = findLCAUtil(root.left, n1, n2) 
    right_lca = findLCAUtil(root.right, n1, n2) 
  
    # If both of the above calls return Non-None, then one node 
    # is present in one subtree and other is present in other, 
    # So this current node is the LCA 
    if (left_lca and right_lca): 
        return root 
  
    # Otherwise check if left subtree or right subtree is LCA 
    return left_lca if (left_lca != None) else right_lca 
  
# Returns True if key k is present in tree rooted with root 
def find(root: Node, k: int) -> bool: 
  
    # Base Case 
    if (root == None): 
        return False
  
    # If key k is present at root, or in left subtree 
    # or right subtree, return True 
    if (root.data == k or find(root.left, k) or find(root.right, k)): 
        return True
  
    # Else return False 
    return False
  
# This function returns LCA of n1 and n2 only if both n1 and n2 
# are present in tree, otherwise returns None 
def findLCA(root: Node, n1: int, n2: int) -> Node: 
    global v1, v2 
  
    # Initialize n1 and n2 as not visited 
    v1 = False
    v2 = False
  
    # Find lca of n1 and n2 
    lca = findLCAUtil(root, n1, n2) 
  
    # Return LCA only if both n1 and n2 are present in tree 
    if (v1 and v2 or v1 and find(lca, n2) or v2 and find(lca, n1)): 
        return lca 
  
    # Else return None 
    return None
  
# function returns True if 
# there is a path from root to 
# the given node. It also populates 
# 'arr' with the given path 
def hasPath(root: Node, arr: list, x: int) -> Node: 
  
    # if root is None 
    # there is no path 
    if (root is None): 
        return False
  
    # push the node's value in 'arr' 
    arr.append(root.data) 
  
    # if it is the required node 
    # return True 
    if (root.data == x): 
        return True
  
    # else check whether there    the required node lies in the 
    # left subtree or right subtree of the current node 
    if (hasPath(root.left, arr, x) or hasPath(root.right, arr, x)): 
        return True
  
    # required node does not lie either in the 
    # left or right subtree of the current node 
    # Thus, remove current node's value from 'arr' 
    # and then return False; 
    arr.pop() 
    return False
  
# function to print the path common 
# to the two paths from the root 
# to the two given nodes if the nodes 
# lie in the binary tree 
def printCommonPath(root: Node, n1: int, n2: int): 
  
    # vector to store the common path 
    arr = [] 
  
    # LCA of node n1 and n2 
    lca = findLCA(root, n1, n2) 
  
    # if LCA of both n1 and n2 exists 
    if (lca): 
  
        # then print the path from root to 
        # LCA node 
        if (hasPath(root, arr, lca.data)): 
  
            for i in range(len(arr) - 1): 
                print(arr[i], end="->") 
            print(arr[-1]) 
  
    # LCA is not present in the binary tree 
    # either n1 or n2 or both are not present 
    else: 
        print("No Common Path") 
  
# Driver Code 
if __name__ == "__main__": 
  
    v1 = 0
    v2 = 0
  
    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.left.right.left = Node(8) 
    root.right.left.right = Node(9) 
  
    n1 = 4
    n2 = 8
    printCommonPath(root, n1, n2) 
  
# This code is contributed by 
# sanjeev2552 

C#

// C# implementation to print the path common to the  
// two paths from the root to the two given nodes  
using System; 
using System.Collections.Generic; 
  
public class PrintCommonPath  
{ 
  
    // Initialize n1 and n2 as not visited  
    static Boolean v1 = false, v2 = false;  
  
    // This function returns pointer to LCA of two given  
    // values n1 and n2. This function assumes that n1 and  
    // n2 are present in Binary Tree  
    static Node findLCAUtil(Node node, int n1, int n2)  
    {  
        // Base case  
        if (node == null)  
            return null;  
              
        //Store result in temp, in case of key 
        // match so that we can search for other key also.  
        Node temp=null;  
      
        // If either n1 or n2 matches with root's key, report the presence  
        // by setting v1 or v2 as true and return root (Note that if a key  
        // is ancestor of other, then the ancestor key becomes LCA)  
        if (node.data == n1)  
        {  
            v1 = true;  
            temp = node;  
        }  
        if (node.data == n2)  
        {  
            v2 = true;  
            temp = node;  
        }  
      
        // Look for keys in left and right subtrees  
        Node left_lca = findLCAUtil(node.left, n1, n2);  
        Node right_lca = findLCAUtil(node.right, n1, n2);  
      
        if (temp != null)  
            return temp;  
      
        // 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 node;  
      
        // Otherwise check if left subtree or right subtree is LCA  
        return (left_lca != null) ? left_lca : right_lca;  
    } 
  
    // Returns true if key k is present in tree rooted with root  
    static Boolean find(Node root, int k)  
    {  
        // Base Case  
        if (root == null)  
            return false;  
      
        // If key k is present at root, or in left subtree  
        // or right subtree, return true  
        if (root.data == k || find(root.left, k) || find(root.right, k))  
            return true;  
      
        // Else return false  
        return false;  
    }  
  
    // This function returns LCA of n1 and n2 only if both n1 and n2  
    // are present in tree, otherwise returns null  
    static Node findLCA(Node root, int n1, int n2)  
    {  
        // Find lca of n1 and n2  
        Node lca = findLCAUtil(root, n1, n2);  
      
        // Return LCA only if both n1 and n2 are present in tree  
        if (v1 && v2 || v1 && find(lca, n2) || v2 && find(lca, n1))  
            return lca;  
      
        // Else return null  
        return null;  
    }  
  
    // function returns true if  
    // there is a path from root to  
    // the given node. It also populates  
    // 'arr' with the given path  
    static Boolean hasPath(Node root, List<int> arr, int x)  
    {  
        // if root is null  
        // there is no path  
        if (root == null)  
            return false;  
          
        // push the node's value in 'arr'  
        arr.Add(root.data);      
          
        // if it is the required node  
        // return true  
        if (root.data == x)      
            return true;  
          
        // else check whether there the required node lies in the  
        // left subtree or right subtree of the current node  
        if (hasPath(root.left, arr, x) ||  
            hasPath(root.right, arr, x))  
            return true;  
          
        // required node does not lie either in the  
        // left or right subtree of the current node  
        // Thus, remove current node's value from 'arr'  
        // and then return false;      
        arr.Remove(arr.Count-1);  
        return false;              
    }  
  
    // function to print the path common  
    // to the two paths from the root  
    // to the two given nodes if the nodes  
    // lie in the binary tree  
    static void printCommonPath(Node root, int n1, int n2)  
    {  
        // ArrayList to store the common path  
        List<int> arr = new List<int>(); 
          
        // LCA of node n1 and n2  
        Node lca = findLCA(root, n1, n2);  
          
        // if LCA of both n1 and n2 exists  
        if (lca!=null)  
        {  
            // then print the path from root to  
            // LCA node  
            if (hasPath(root, arr, lca.data))  
            {  
                for (int i=0; i<arr.Count-1; i++)      
                    Console.Write(arr[i]+"->"); 
                    Console.Write(arr[arr.Count - 1]);      
            }      
        }  
          
        // LCA is not present in the binary tree  
        // either n1 or n2 or both are not present  
        else
            Console.Write("No Common Path"); 
    }  
      
    // 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.right.left = new Node(6);  
        root.right.right = new Node(7);  
        root.left.right.left = new Node(8);  
        root.right.left.right = new Node(9);  
              
        int n1 = 4, n2 = 8;  
        printCommonPath(root, n1, n2);  
        } 
} 
  
/* Class containing left and right child of current  
node and key value*/
public class Node  
{  
    public int data;  
    public Node left, right;  
      
    public Node(int item)  
    {  
        data = item;  
        left = right = null;  
    }  
}  
  
// This code has been contributed by 29AjayKumar 

Output:

1->2

Time complexity: O(n), where n is the number of nodes in the binary tree.

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