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Check if all elements of given Linked List corresponds to a downward path from any node in given Binary Tree

  • Difficulty Level : Basic
  • Last Updated : 22 Nov, 2021

Given a root of the Binary Tree and head of the Linked List, the task is to check if all the elements of the linked list corresponds to a downward path from any node in the given Binary Tree.

Examples:

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Input: Tree in the image below, list = {3, 6, 8}



Output: Yes
Explanation: There exists a downward path in the given Binary Tree, having all the elements of the linked list in the same order.

Input: Tree in the image below, list = {1, 2, 5, 7}
 

Output: Yes

Approach: The given problem can be solved with the help of the DFS Traversal of the Binary tree. During the DFS traversal, if any node of the Binary Tree is equal to the head of the linked list, a recursive function isPathUntil() can be called to recursively check whether the other elements of the linked list also exist as a path from that node. If the complete linked list has been traversed, there exists a valid required path, hence return true. Otherwise, return false. Follow the steps below to solve the given problem:

  • Declare a function, say isSubPathUtil(root, head) , and perform the following steps in this function:
    • If the root is NULL, then return false.
    • If the head is NULL, then return true.
    • If the value of the current root node is the same as the value of the current head of LL, then recursively call for isSubPathUtil(root->left, head->next) and isSubPathUtil(root->right, head->next) and if the value returned one of these recursive calls is true, then return true. Otherwise, return false.
  • Declare a function, say isSubPath(root, head), and perform the following steps in this function:
    • If the root is NULL, then return false.
    • If the head is NULL, then return true.
    • If the value of the current root node is the same as the value of the current head of LL, then recursively call for isSubPath(root->left, head->next) and isSubPath(root->right, head->next) and if the value returned one of these recursive calls is true, then return true. Otherwise, return value call recursively for isSubPath(root->left, head) and isSubPath(root->right, head).
  • After the above steps, if the value returned by the function isSubPath(root, head) is true , then print Yes . Otherwise, print No .

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include "bits/stdc++.h"
using namespace std;
 
// Node of the Linked list
struct listNode {
    int val;
    struct listNode* next;
 
    // Constructor
    listNode(int data)
    {
        this->val = data;
        next = NULL;
    }
};
 
// Node of the Binary Search tree
struct treeNode {
    int val;
    treeNode* right;
    treeNode* left;
 
    // Constructor
    treeNode(int data)
    {
        this->val = data;
        this->left = NULL;
        this->right = NULL;
    }
};
 
// Recursive function to check if there
// exist a path from the node curTree
// having the LL from the node curList
bool isPathUtil(listNode* curList,
                treeNode* curTree)
{
    // If the complete linked list
    // is traversed
    if (curList == NULL)
        return true;
 
    // If the tree node doesnot exist
    if (curTree == NULL)
        return false;
 
    if (curList->val == curTree->val) {
 
        // Recursively calling for the
        // next element
        return isPathUtil(curList->next,
                          curTree->left)
               || isPathUtil(curList->next,
                             curTree->right);
    }
    else {
 
        // If not found, return false
        return false;
    }
}
 
// Function to check if the linked list
// exist as a downward path in Binary tree
// using the DFS Traversal of the Tree
bool isPath(listNode* head, treeNode* root)
{
 
    // If the current node of the
    // tree is Null
    if (root == NULL)
        return false;
 
    // If the complete linked list
    // has been found
    if (head == NULL)
        return true;
 
    // Stores if there exist the
    // required path
    bool isPossible = false;
 
    if (root->val == head->val) {
 
        // Recursively calling to
        // check the next node of
        // the linked list
        isPossible = isPathUtil(
                         head->next, root->left)
                     || isPathUtil(
                            head->next, root->right);
    }
 
    // Recursive calling for the next
    // elements of of the binary tree
    return isPossible || isPath(head, root->left)
           || isPath(head, root->right);
}
 
// Driver Code
int main()
{
    treeNode* root = new treeNode(1);
    root->left = new treeNode(2);
    root->right = new treeNode(3);
    root->left->left = new treeNode(4);
    root->left->right = new treeNode(5);
    root->left->right->left = new treeNode(7);
    root->right->right = new treeNode(6);
    root->right->right->left = new treeNode(8);
    root->right->right->right = new treeNode(9);
 
    listNode* head = new listNode(3);
    head->next = new listNode(6);
    head->next->next = new listNode(8);
 
    // Function Call
    cout << (isPath(head, root) ? "Yes" : "No");
 
    return 0;
}

Java




// Java program for the above approach
 
//include "bits/stdJava.h"
import java.util.*;
 
class GFG{
 
// Node of the Linked list
static class listNode {
    int val;
    listNode next;
 
    // Constructor
    listNode(int data)
    {
        this.val = data;
        next = null;
    }
};
 
// Node of the Binary Search tree
static class treeNode {
    int val;
    treeNode right;
    treeNode left;
 
    // Constructor
    treeNode(int data)
    {
        this.val = data;
        this.left = null;
        this.right = null;
    }
};
 
// Recursive function to check if there
// exist a path from the node curTree
// having the LL from the node curList
static boolean isPathUtil(listNode curList,
                treeNode curTree)
{
    // If the complete linked list
    // is traversed
    if (curList == null)
        return true;
 
    // If the tree node doesnot exist
    if (curTree == null)
        return false;
 
    if (curList.val == curTree.val) {
 
        // Recursively calling for the
        // next element
        return isPathUtil(curList.next,
                          curTree.left)
               || isPathUtil(curList.next,
                             curTree.right);
    }
    else {
 
        // If not found, return false
        return false;
    }
}
 
// Function to check if the linked list
// exist as a downward path in Binary tree
// using the DFS Traversal of the Tree
static boolean isPath(listNode head, treeNode root)
{
 
    // If the current node of the
    // tree is Null
    if (root == null)
        return false;
 
    // If the complete linked list
    // has been found
    if (head == null)
        return true;
 
    // Stores if there exist the
    // required path
    boolean isPossible = false;
 
    if (root.val == head.val) {
 
        // Recursively calling to
        // check the next node of
        // the linked list
        isPossible = isPathUtil(
                         head.next, root.left)
                     || isPathUtil(
                            head.next, root.right);
    }
 
    // Recursive calling for the next
    // elements of of the binary tree
    return isPossible || isPath(head, root.left)
           || isPath(head, root.right);
}
 
// Driver Code
public static void main(String[] args)
{
    treeNode root = new treeNode(1);
    root.left = new treeNode(2);
    root.right = new treeNode(3);
    root.left.left = new treeNode(4);
    root.left.right = new treeNode(5);
    root.left.right.left = new treeNode(7);
    root.right.right = new treeNode(6);
    root.right.right.left = new treeNode(8);
    root.right.right.right = new treeNode(9);
 
    listNode head = new listNode(3);
    head.next = new listNode(6);
    head.next.next = new listNode(8);
 
    // Function Call
    System.out.print((isPath(head, root) ? "Yes" : "No"));
 
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python program for the above approach
 
# Node of the Linked list
class listNode:
 
  # Constructor
  def __init__ (self, data):
    self.val = data;
    self.next = None;
 
# Node of the Binary Search tree
class treeNode:
 
  # Constructor
  def __init__ (self, data):
    self.val = data;
    self.left = None;
    self.right = None;
   
# Recursive function to check if there
# exist a path from the node curTree
# having the LL from the node curList
def isPathUtil(curList, curTree):
 
  # If the complete linked list
  # is traversed
  if (curList == None): return True;
 
  # If the tree node doesnot exist
  if (curTree == None): return False;
 
  if (curList.val == curTree.val):
   
    # Recursively calling for the
    # next element
    return (
      isPathUtil(curList.next, curTree.left) or
      isPathUtil(curList.next, curTree.right)
    );
   
  else:
   
    # If not found, return false
    return False;
   
# Function to check if the linked list
# exist as a downward path in Binary tree
# using the DFS Traversal of the Tree
def isPath(head, root):
 
  # If the current node of the
  # tree is None
  if (root == None): return False;
 
  # If the complete linked list
  # has been found
  if (head == None): return True;
 
  # Stores if there exist the
  # required path
  isPossible = False;
 
  if (root.val == head.val):
    # Recursively calling to
    # check the next node of
    # the linked list
    isPossible = isPathUtil(head.next, root.left) or isPathUtil(head.next, root.right);
   
 
  # Recursive calling for the next
  # elements of of the binary tree
  return isPossible or isPath(head, root.left) or isPath(head, root.right);
 
 
# Driver Code
 
root = treeNode(1);
root.left = treeNode(2);
root.right = treeNode(3);
root.left.left = treeNode(4);
root.left.right = treeNode(5);
root.left.right.left = treeNode(7);
root.right.right = treeNode(6);
root.right.right.left = treeNode(8);
root.right.right.right = treeNode(9);
 
head = listNode(3);
head.next = listNode(6);
head.next.next = listNode(8);
 
# Function Call
print( "Yes" if isPath(head, root) else "No");
 
# This code is contributed by saurabh_jaiswal.

C#




// C# program for the above approach
 
//include "bits/stdJava.h"
using System;
 
public class GFG{
 
// Node of the Linked list
class listNode {
    public int val;
    public listNode next;
 
    // Constructor
    public listNode(int data)
    {
        this.val = data;
        next = null;
    }
};
 
// Node of the Binary Search tree
class treeNode {
    public int val;
    public treeNode right;
    public treeNode left;
 
    // Constructor
    public treeNode(int data)
    {
        this.val = data;
        this.left = null;
        this.right = null;
    }
};
 
// Recursive function to check if there
// exist a path from the node curTree
// having the LL from the node curList
static bool isPathUtil(listNode curList,
                treeNode curTree)
{
    // If the complete linked list
    // is traversed
    if (curList == null)
        return true;
 
    // If the tree node doesnot exist
    if (curTree == null)
        return false;
 
    if (curList.val == curTree.val) {
 
        // Recursively calling for the
        // next element
        return isPathUtil(curList.next,
                          curTree.left)
               || isPathUtil(curList.next,
                             curTree.right);
    }
    else {
 
        // If not found, return false
        return false;
    }
}
 
// Function to check if the linked list
// exist as a downward path in Binary tree
// using the DFS Traversal of the Tree
static bool isPath(listNode head, treeNode root)
{
 
    // If the current node of the
    // tree is Null
    if (root == null)
        return false;
 
    // If the complete linked list
    // has been found
    if (head == null)
        return true;
 
    // Stores if there exist the
    // required path
    bool isPossible = false;
 
    if (root.val == head.val) {
 
        // Recursively calling to
        // check the next node of
        // the linked list
        isPossible = isPathUtil(
                         head.next, root.left)
                     || isPathUtil(
                            head.next, root.right);
    }
 
    // Recursive calling for the next
    // elements of of the binary tree
    return isPossible || isPath(head, root.left)
           || isPath(head, root.right);
}
 
// Driver Code
public static void Main(String[] args)
{
    treeNode root = new treeNode(1);
    root.left = new treeNode(2);
    root.right = new treeNode(3);
    root.left.left = new treeNode(4);
    root.left.right = new treeNode(5);
    root.left.right.left = new treeNode(7);
    root.right.right = new treeNode(6);
    root.right.right.left = new treeNode(8);
    root.right.right.right = new treeNode(9);
 
    listNode head = new listNode(3);
    head.next = new listNode(6);
    head.next.next = new listNode(8);
 
    // Function Call
    Console.Write((isPath(head, root) ? "Yes" : "No"));
 
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
// Javascript program for the above approach
 
// Node of the Linked list
class listNode
{
 
  // Constructor
  constructor(data) {
    this.val = data;
    this.next = null;
  }
}
 
// Node of the Binary Search tree
class treeNode
{
 
  // Constructor
  constructor(data) {
    this.val = data;
    this.left = null;
    this.right = null;
  }
}
 
// Recursive function to check if there
// exist a path from the node curTree
// having the LL from the node curList
function isPathUtil(curList, curTree)
{
 
  // If the complete linked list
  // is traversed
  if (curList == null) return true;
 
  // If the tree node doesnot exist
  if (curTree == null) return false;
 
  if (curList.val == curTree.val)
  {
   
    // Recursively calling for the
    // next element
    return (
      isPathUtil(curList.next, curTree.left) ||
      isPathUtil(curList.next, curTree.right)
    );
  }
  else
  {
   
    // If not found, return false
    return false;
  }
}
 
// Function to check if the linked list
// exist as a downward path in Binary tree
// using the DFS Traversal of the Tree
function isPath(head, root)
{
 
  // If the current node of the
  // tree is null
  if (root == null) return false;
 
  // If the complete linked list
  // has been found
  if (head == null) return true;
 
  // Stores if there exist the
  // required path
  let isPossible = false;
 
  if (root.val == head.val) {
    // Recursively calling to
    // check the next node of
    // the linked list
    isPossible =
      isPathUtil(head.next, root.left) || isPathUtil(head.next, root.right);
  }
 
  // Recursive calling for the next
  // elements of of the binary tree
  return isPossible || isPath(head, root.left) || isPath(head, root.right);
}
 
// Driver Code
 
let root = new treeNode(1);
root.left = new treeNode(2);
root.right = new treeNode(3);
root.left.left = new treeNode(4);
root.left.right = new treeNode(5);
root.left.right.left = new treeNode(7);
root.right.right = new treeNode(6);
root.right.right.left = new treeNode(8);
root.right.right.right = new treeNode(9);
 
let head = new listNode(3);
head.next = new listNode(6);
head.next.next = new listNode(8);
 
// Function Call
document.write(isPath(head, root) ? "Yes" : "No");
 
// This code is contributed by saurabh_jaiswal.
</script>
Output: 
Yes

 

Time Complexity: O(N * M) where N = Number of nodes in the Binary Tree and M = Number of nodes in the Linked list.
Auxiliary Space: O(1)




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