Given a binary tree and an array, the task is to find if the given array sequence is present as a root to leaf path in given tree.

Examples:

Input : arr[] = {5, 8, 6, 7} & above tree
Output: "Path Exist"
Input :  arr[] = {5, 3, 4, 9} & above tree
Output: "Path does not Exist"
Input : {5, 6, 2} & above tree
Output: "Path does not Exist"

A simple solution for this problem is to find all root to leaf paths in given tree and for each root to leaf path check that path and given sequence in array both are identical or not.

An efficient solution for this problem is to traverse the tree once and while traversing the tree we have to check that if path from root to current node is identical to the given sequence of root to leaf path. Here is the algorithm :

• Start traversing tree in preorder fashion.
• Whenever we moves down in tree then we also move by one index in given sequence of root to leaf path .
• If current node is equal to the arr[index] this means that till this level of tree path is identical.
• Now remaining path will either be in left subtree or in right subtree.
• If any node gets mismatched with arr[index] this means that current path is not identical to the given sequence of root to leaf path, so we return back and move in right subtree.
• Now when we are at leaf node and it is equal to arr[index] and there is no further element in given sequence of root to leaf path, this means that path exist in given tree.

// C++ program to see if there is a root to leaf path
// with given sequence.
#include<bits/stdc++.h>
using namespace std;
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct Node
{
    int data;
    struct Node* left, *right;
};
/* utility that allocates a new node with the
given data and NULL left and right pointers. */
struct Node* newnode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right  = NULL;
    return (node);
}
// function to check given sequence of root to leaf path exist
// in tree or not.
// index represents current element in sequence of rooth to
// leaf path
bool existPath(struct Node *root, int arr[], int n, int index)
{
    // If root is NULL, then there must not be any element
    // in array.
    if (root == NULL)
        return (n == 0);
   // If this node is a leaf and matches with last entry
   // of array.
   if ((root->left == NULL && root->right == NULL) &&
       (root->data == arr[index]) && (index == n-1))
            return true;
   // If current node is equal to arr[index] this means
   // that till this level path has been matched and
   // remaining path can be either in left subtree or
   // right subtree.
   return ((index < n) && (root->data == arr[index]) &&
              (existPath(root->left, arr, n,  index+1) ||
               existPath(root->right, arr, n, index+1) ));
}
// Driver function to run the case
int main()
{
    // arr[] --> sequence of root to leaf path
    int arr[] = {5, 8, 6, 7};
    int n = sizeof(arr)/sizeof(arr[0]);
    struct Node *root = newnode(5);
    root->left    = newnode(3);
    root->right   = newnode(8);
    root->left->left = newnode(2);
    root->left->right = newnode(4);
    root->left->left->left = newnode(1);
    root->right->left = newnode(6);
    root->right->left->right = newnode(7);
    existPath(root, arr, n, 0)? cout << "Path Exists" :
                                cout << "Path does not Exist";
    return 0;
}

Output:

Path Exists

Time complexity : O(n)