# Check if two nodes are on same path in a tree | Set 2

Given two nodes of a binary tree v1 and v2, the task is to check if two nodes are on the same path in a tree.

Example:

```Input:  v1 = 1, v2 = 5
1
/  |  \
2   3   4
/    |    \
5     6     7

Output: Yes
Explanation:
Both nodes 1 and 5
lie in the path 1 -> 2 -> 5.

Input: v1 = 2, v2 = 6
1
/  |  \
2   3   4
/    |    \
5     6     7

Output: NO
```

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

DFS Approach: Refer to Check if two nodes are on same path in a tree for the DFS approach.

LCA Approach: The idea is to use Lowest Common Ancestor. Find the LCA of the given vertices v1 and v2. If the LCA is equal to any of the given two vertices, print Yes. Otherwise, print No.

Below is the implementation of above approach:

## C++

 `// C++ program to check if two nodes ` `// are on same path in a tree without ` `// using any extra space ` `#include ` `using` `namespace` `std; ` ` `  `// Function to filter ` `// the return Values ` `int` `filter(``int` `x, ``int` `y, ``int` `z) ` `{ ` `    ``if` `(x != -1 && y != -1) { ` `        ``return` `z; ` `    ``} ` `    ``return` `x == -1 ? y : x; ` `} ` ` `  `// Utility function to check if nodes ` `// are on same path or not ` `int` `samePathUtil(``int` `mtrx[], ``int` `vrtx, ` `                 ``int` `v1, ``int` `v2, ``int` `i) ` `{ ` `    ``int` `ans = -1; ` ` `  `    ``// Condition to check ` `    ``// if any vertex ` `    ``// is equal to given two ` `    ``// vertex or not ` `    ``if` `(i == v1 || i == v2) ` `        ``return` `i; ` ` `  `    ``for` `(``int` `j = 0; j < vrtx; j++) { ` ` `  `        ``// Check if the current ` `        ``// position has 1 ` `        ``if` `(mtrx[i][j] == 1) { ` `            ``// Recursive call ` `            ``ans ` `                ``= filter(ans, ` `                         ``samePathUtil(mtrx, ` `                                      ``vrtx, ` `                                      ``v1, ` `                                      ``v2, ` `                                      ``j), ` `                         ``i); ` `        ``} ` `    ``} ` ` `  `    ``// Return LCA ` `    ``return` `ans; ` `} ` ` `  `// Function to check if nodes ` `// lies on same path or not ` `bool` `isVertexAtSamePath(``int` `mtrx[], ` `                        ``int` `vrtx, ``int` `v1, ` `                        ``int` `v2, ``int` `i) ` `{ ` `    ``int` `lca = samePathUtil(mtrx, ` `                           ``vrtx, v1 - 1, ` `                           ``v2 - 1, i); ` ` `  `    ``if` `(lca == v1 - 1 || lca == v2 - 1) ` `        ``return` `true``; ` ` `  `    ``return` `false``; ` `} ` ` `  `// Driver Program ` `int` `main() ` `{ ` `    ``int` `vrtx = 7, edge = 6; ` `    ``int` `mtrx = { ` `        ``{ 0, 1, 1, 1, 0, 0, 0 }, ` `        ``{ 0, 0, 0, 0, 1, 0, 0 }, ` `        ``{ 0, 0, 0, 0, 0, 1, 0 }, ` `        ``{ 0, 0, 0, 0, 0, 0, 1 }, ` `        ``{ 0, 0, 0, 0, 0, 0, 0 }, ` `        ``{ 0, 0, 0, 0, 0, 0, 0 }, ` `        ``{ 0, 0, 0, 0, 0, 0, 0 } ` ` `  `    ``}; ` ` `  `    ``int` `v1 = 1, v2 = 5; ` ` `  `    ``if` `(isVertexAtSamePath(mtrx, ` `                           ``vrtx, v1, ` `                           ``v2, 0)) ` `        ``cout << ``"Yes"``; ` ` `  `    ``else` `        ``cout << ``"No"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to check if two nodes ` `// are on same path in a tree without ` `// using any extra space ` `class` `GFG{ ` ` `  `// Function to filter ` `// the return Values ` `static` `int` `filter(``int` `x, ``int` `y, ``int` `z) ` `{ ` `    ``if` `(x != -``1` `&& y != -``1``) ` `    ``{ ` `        ``return` `z; ` `    ``} ` `    ``return` `x == -``1` `? y : x; ` `} ` ` `  `// Utility function to check if nodes ` `// are on same path or not ` `static` `int` `samePathUtil(``int` `mtrx[][], ``int` `vrtx, ` `                        ``int` `v1, ``int` `v2, ``int` `i) ` `{ ` `    ``int` `ans = -``1``; ` ` `  `    ``// Condition to check ` `    ``// if any vertex ` `    ``// is equal to given two ` `    ``// vertex or not ` `    ``if` `(i == v1 || i == v2) ` `        ``return` `i; ` ` `  `    ``for``(``int` `j = ``0``; j < vrtx; j++) ` `    ``{ ` `         `  `        ``// Check if the current ` `        ``// position has 1 ` `        ``if` `(mtrx[i][j] == ``1``) ` `        ``{ ` `             `  `            ``// Recursive call ` `            ``ans = filter(ans, samePathUtil( ` `                         ``mtrx, vrtx, v1, ` `                         ``v2, j), i); ` `        ``} ` `    ``} ` ` `  `    ``// Return LCA ` `    ``return` `ans; ` `} ` ` `  `// Function to check if nodes ` `// lies on same path or not ` `static` `boolean` `isVertexAtSamePath(``int` `mtrx[][], ` `                                  ``int` `vrtx, ``int` `v1, ` `                                  ``int` `v2, ``int` `i) ` `{ ` `    ``int` `lca = samePathUtil(mtrx, vrtx, v1 - ``1``, ` `                                       ``v2 - ``1``, i); ` `                                        `  `    ``if` `(lca == v1 - ``1` `|| lca == v2 - ``1``) ` `        ``return` `true``; ` ` `  `    ``return` `false``; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `vrtx = ``7``; ` `    ``int` `mtrx[][] = { { ``0``, ``1``, ``1``, ``1``, ``0``, ``0``, ``0` `}, ` `                     ``{ ``0``, ``0``, ``0``, ``0``, ``1``, ``0``, ``0` `}, ` `                     ``{ ``0``, ``0``, ``0``, ``0``, ``0``, ``1``, ``0` `}, ` `                     ``{ ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``1` `}, ` `                     ``{ ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0` `}, ` `                     ``{ ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0` `}, ` `                     ``{ ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0` `} }; ` ` `  `    ``int` `v1 = ``1``, v2 = ``5``; ` ` `  `    ``if` `(isVertexAtSamePath(mtrx, vrtx, ` `                           ``v1, v2, ``0``)) ` `        ``System.out.print(``"Yes"``); ` `    ``else` `        ``System.out.print(``"No"``); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```Yes
```

Time Complexity: O(N)
Auxiliary Space: O(1) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : Rajput-Ji