# Check if there is a root to leaf path with given sequence

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, 2, 4, 8} for above tree Output: "Path Exist" Input : arr[] = {5, 3, 4, 9} for 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++

`// 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);`

`}`

`// Util function`

`bool`

`existPathUtil(`

`struct`

`Node *root,`

`int`

`arr[],`

`int`

`n,`

`int`

`index)`

`{`

`// If root is NULL or reached end of the array`

`if`

`(root == NULL or index==n)`

`return`

`false`

`;`

`// If current node is leaf`

`if`

`(root->left == NULL && root->right == NULL)`

`{`

`if`

`((root->data == arr[index]) && (index == n-1))`

`return`

`true`

`;`

`return`

`false`

`;`

`}`

`// 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]) &&`

`(existPathUtil(root->left, arr, n, index+1) ||`

`existPathUtil(root->right, arr, n, index+1) ));`

`}`

`// 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)`

`return`

`(n==0);`

`return`

`existPathUtil(root, arr, n, 0);`

`}`

`// 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;`

`}`

## Java

`// Java program to see if there is a root to leaf path`

`// with given sequence.`

`import`

`java.io.*;`

`class`

`Node`

`{`

`int`

`data;`

`Node left;`

`Node right;`

`Node(`

`int`

`data)`

`{`

`this`

`.data = data;`

`this`

`.left =`

`null`

`;`

`this`

`.right =`

`null`

`;`

`}`

`}`

`class`

`GFG {`

`// Util function`

`static`

`boolean`

`existPathUtil(Node root,`

`int`

`arr[],`

`int`

`n,`

`int`

`index)`

`{`

`// If root is NULL or`

`// reached end of the array`

`if`

`(root ==`

`null`

`|| index==n)`

`return`

`false`

`;`

`// If current node is leaf`

`if`

`(root.left ==`

`null`

`&&`

`root.right ==`

`null`

`)`

`{`

`if`

`((root.data == arr[index]) &&`

`(index == n-`

`1`

`))`

`return`

`true`

`;`

`return`

`false`

`;`

`}`

`// 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])`

`&& (existPathUtil(root.left, arr, n, index+`

`1`

`)`

`|| existPathUtil(root.right, arr, n, index+`

`1`

`) ));`

`}`

`// Function to check given sequence of root`

`// to leaf path exist in tree or not.`

`// index : current element in sequence of root to`

`// leaf path`

`static`

`boolean`

`existPath(Node root,`

`int`

`arr[],`

`int`

`n,`

`int`

`index)`

`{`

`if`

`(root ==`

`null`

`)`

`return`

`(n==`

`0`

`);`

`return`

`existPathUtil(root, arr, n,`

`0`

`);`

`}`

`public`

`static`

`void`

`main (String[] args) {`

`// arr[] : sequence of root to leaf path`

`int`

`arr[] = {`

`5`

`,`

`8`

`,`

`6`

`,`

`7`

`};`

`int`

`n = arr.length;`

`Node root =`

`new`

`Node(`

`5`

`);`

`root.left =`

`new`

`Node(`

`3`

`);`

`root.right =`

`new`

`Node(`

`8`

`);`

`root.left.left =`

`new`

`Node(`

`2`

`);`

`root.left.right =`

`new`

`Node(`

`4`

`);`

`root.left.left.left =`

`new`

`Node(`

`1`

`);`

`root.right.left =`

`new`

`Node(`

`6`

`);`

`root.right.left.right =`

`new`

`Node(`

`7`

`);`

`if`

`(existPath(root, arr, n,`

`0`

`))`

`System.out.println(`

`"Path Exists"`

`);`

`else`

`System.out.println(`

`"Path does not Exist"`

`);`

`}`

`}`

## Python3

`# Python program to see if`

`# there is a root to leaf path`

`# with given sequence`

`# Class of Node`

`class`

`Node:`

`# Constructor to create a`

`# node in Binary Tree`

`def`

`__init__(`

`self`

`, val):`

`self`

`.val`

`=`

`val`

`self`

`.left`

`=`

`None`

`self`

`.right`

`=`

`None`

`# Util function`

`def`

`existPathUtil(root, arr, n, index):`

`# If root is NULL or reached`

`# end of the array`

`if`

`not`

`root`

`or`

`index`

`=`

`=`

`n:`

`return`

`False`

`# If current node is leaf`

`if`

`not`

`root.left`

`and`

`not`

`root.right:`

`if`

`root.val`

`=`

`=`

`arr[index]`

`and`

`index`

`=`

`=`

`n`

`-`

`1`

`:`

`return`

`True`

`return`

`False`

`# 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)`

`and`

`(root.val`

`=`

`=`

`arr[index])`

`and`

`\`

`(existPathUtil(root.left, arr, n, index`

`+`

`1`

`)`

`or`

`\`

`existPathUtil(root.right, arr, n, index`

`+`

`1`

`)))`

`# 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`

`def`

`existPath(root, arr, n, index):`

`if`

`not`

`root:`

`return`

`(n`

`=`

`=`

`0`

`)`

`return`

`existPathUtil(root, arr, n,`

`0`

`)`

`# Driver Code`

`if`

`__name__`

`=`

`=`

`"__main__"`

`:`

`arr`

`=`

`[`

`5`

`,`

`8`

`,`

`6`

`,`

`7`

`]`

`n`

`=`

`len`

`(arr)`

`root`

`=`

`Node(`

`5`

`)`

`root.left`

`=`

`Node(`

`3`

`)`

`root.right`

`=`

`Node(`

`8`

`)`

`root.left.left`

`=`

`Node(`

`2`

`)`

`root.left.right`

`=`

`Node(`

`4`

`)`

`root.left.left.left`

`=`

`Node(`

`1`

`)`

`root.right.left`

`=`

`Node(`

`6`

`)`

`root.right.left.right`

`=`

`Node(`

`7`

`)`

`if`

`existPath(root, arr, n,`

`0`

`):`

`print`

`(`

`"Path Exists"`

`)`

`else`

`:`

`print`

`(`

`"Path does not Exist"`

`)`

## C#

`// C# program to see if there`

`// is a root to leaf path`

`// with given sequence.`

`using`

`System;`

`public`

`class`

`CheckForPath`

`{`

`// 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`

`public`

`static`

`bool`

`existPath(Node root,`

`int`

`[]arr,`

`int`

`index)`

`{`

`// If root is NULL, then there`

`// must not be any element`

`// in array.`

`if`

`(root ==`

`null`

`)`

`{`

`return`

`arr.Length == 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] &&`

`root.data == arr[arr.Length - 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 < arr.Length && (root.data == arr[index] &&`

`(existPath(root.left,arr,index + 1) ||`

`existPath(root.right, arr, index + 1))));`

`}`

`// Driver code`

`public`

`static`

`void`

`Main()`

`{`

`// arr[] is sequence of root to leaf path`

`int`

`[]arr = {5, 8, 6, 7};`

`Node root=`

`new`

`Node(5);`

`root.left=`

`new`

`Node(3);`

`root.right=`

`new`

`Node(8);`

`root.left.left =`

`new`

`Node(2);`

`root.left.right =`

`new`

`Node(4);`

`root.left.left.left =`

`new`

`Node(1);`

`root.right.left =`

`new`

`Node(6);`

`root.right.left.right =`

`new`

`Node(7);`

`if`

`(existPath(root, arr, 0))`

`{`

`Console.Write(`

`"Path Exists"`

`);`

`}`

`else`

`{`

`Console.Write(`

`"Path does not Exist"`

`);`

`}`

`}`

`}`

`/* A binary tree node has data,`

`pointer to left child and a`

`pointer to right child */`

`public`

`class`

`Node`

`{`

`public`

`int`

`data;`

`public`

`Node left, right;`

`public`

`Node(`

`int`

`data)`

`{`

`this`

`.data = data;`

`left = right =`

`null`

`;`

`}`

`};`

`// This code is contributed Rajput-Ji`

**Output:**Path Exists

**Time complexity :**O(n)This article is contributed by

**Shashank Mishra ( Gullu )**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.