# Print path between any two nodes in a Binary Tree | Set 2

Given a Binary Tree of distinct nodes and a pair of nodes. The task is to find and print the path between the two given nodes in the binary tree.

Examples:

Input: N1 = 7, N2 = 4 Output: 7 3 1 4

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

Approach: An approach to solve this problem has been discussed in this article. In this article, an even optimised recursive approach will be discussed.
In this recursive approach, below are the steps:

1. Find the first value recursively, once found add the value to the stack.
2. Now every node that is visited whether in backtracking or forward tracking, add the values to the stack but if the node was added in the forward tracking then remove it in the backtracking and continue this until the second value is found or all nodes are visited.

For example: Consider the path between 7 and 9 is to be found in the above tree. We traverse the tree as DFS, once we find the value 7, we add it to the stack. Traversing path 0 -> 1 -> 3 -> 7.
Now while backtracking, add 3 and 1 to the stack. So as of now, the stack has [7, 3, 1], child 1 has a right child so we first add it to the stack. Now, the stack contains [7, 3, 1, 4]. Visit the left child of 4, add it to the stack. The stack contains [7, 3, 1, 4, 8] now. Since there is no further node we would go back to the previous node and since 8 was already added to the stack so remove it. Now the node 4 has a right child and we add it to the stack since this is the second value we were looking for there won’t be any further recursive calls. Finally, the stack contains [7, 3, 1, 4, 9].

Below is the implementation of the above approach:

 `// Java implementation of the approach ` `import` `java.util.Stack; ` ` `  `public` `class` `GFG { ` ` `  `    ``// A binary tree node ` `    ``private` `static` `class` `Node { ` `        ``public` `Node left; ` `        ``public` `int` `value; ` `        ``public` `Node right; ` ` `  `        ``public` `Node(``int` `value) ` `        ``{ ` `            ``this``.value = value; ` `            ``left = ``null``; ` `            ``right = ``null``; ` `        ``} ` `    ``} ` ` `  `    ``private` `boolean` `firstValueFound = ``false``; ` `    ``private` `boolean` `secondValueFound = ``false``; ` `    ``private` `Stack stack = ``new` `Stack(); ` `    ``private` `Node root = ``null``; ` ` `  `    ``public` `GFG(Node root) ` `    ``{ ` `        ``this``.root = root; ` `    ``} ` ` `  `    ``// Function to find the path between ` `    ``// two nodes in binary tree ` `    ``public` `Stack pathBetweenNode(``int` `v1, ``int` `v2) ` `    ``{ ` `        ``pathBetweenNode(``this``.root, v1, v2); ` ` `  `        ``// If both the values are found ` `        ``// then return the stack ` `        ``if` `(firstValueFound && secondValueFound) { ` `            ``return` `stack; ` `        ``} ` ` `  `        ``// If none of the two values is ` `        ``// found then return empty stack ` `        ``return` `new` `Stack(); ` `    ``} ` ` `  `    ``// Function to find the path between ` `    ``// two nodes in binary tree ` `    ``private` `void` `pathBetweenNode(Node root, ``int` `v1, ``int` `v2) ` `    ``{ ` `        ``// Base condition ` `        ``if` `(root == ``null``) ` `            ``return``; ` ` `  `        ``// If both the values are found then return ` `        ``if` `(firstValueFound && secondValueFound) ` `            ``return``; ` ` `  `        ``// Starting the stack frame with ` `        ``// isAddedToStack = false flag ` `        ``boolean` `isAddedToStack = ``false``; ` ` `  `        ``// If one of the value is found then add the ` `        ``// value to the stack and make the isAddedToStack = true ` `        ``if` `(firstValueFound ^ secondValueFound) { ` `            ``stack.add(root); ` `            ``isAddedToStack = ``true``; ` `        ``} ` ` `  `        ``// If none of the two values is found ` `        ``if` `(!(firstValueFound && secondValueFound)) { ` `            ``pathBetweenNode(root.left, v1, v2); ` `        ``} ` ` `  `        ``// Ccopy of current state of firstValueFound ` `        ``// and secondValueFound flag ` `        ``boolean` `localFirstValueFound = firstValueFound; ` `        ``boolean` `localSecondValueFound = secondValueFound; ` ` `  `        ``// If the first value is found ` `        ``if` `(root.value == v1) ` `            ``firstValueFound = ``true``; ` ` `  `        ``// If the second value is found ` `        ``if` `(root.value == v2) ` `            ``secondValueFound = ``true``; ` ` `  `        ``boolean` `localAdded = ``false``; ` ` `  `        ``// If one of the value is found and the value ` `        ``// was not added to the stack yet or there was ` `        ``// only one value found and now both the values ` `        ``// are found and was not added to ` `        ``// the stack then add it ` `        ``if` `(((firstValueFound ^ secondValueFound) ` `             ``|| ((localFirstValueFound ^ localSecondValueFound) ` `                 ``&& (firstValueFound && secondValueFound))) ` `            ``&& !isAddedToStack) { ` `            ``localAdded = ``true``; ` `            ``stack.add(root); ` `        ``} ` ` `  `        ``// If none of the two values is found yet ` `        ``if` `(!(firstValueFound && secondValueFound)) { ` `            ``pathBetweenNode(root.right, v1, v2); ` `        ``} ` ` `  `        ``if` `((firstValueFound ^ secondValueFound) ` `            ``&& !isAddedToStack && !localAdded) ` `            ``stack.add(root); ` ` `  `        ``if` `((firstValueFound ^ secondValueFound) ` `            ``&& isAddedToStack) ` `            ``stack.pop(); ` `    ``} ` ` `  `    ``// Recursive function to print the ` `    ``// contents of a stack in reverse ` `    ``private` `static` `void` `print(Stack stack) ` `    ``{ ` ` `  `        ``// If the stack is empty ` `        ``if` `(stack.isEmpty()) ` `            ``return``; ` ` `  `        ``// Get the top value ` `        ``int` `value = stack.pop().value; ` ` `  `        ``// Recursive call ` `        ``print(stack); ` ` `  `        ``// Print the popped value ` `        ``System.out.print(value + ``" "``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``Node root = ``new` `Node(``0``); ` `        ``root.left = ``new` `Node(``1``); ` `        ``root.right = ``new` `Node(``2``); ` `        ``root.left.left = ``new` `Node(``3``); ` `        ``root.left.right = ``new` `Node(``4``); ` `        ``root.right.left = ``new` `Node(``5``); ` `        ``root.right.right = ``new` `Node(``6``); ` `        ``root.left.left.left = ``new` `Node(``7``); ` `        ``root.left.right.left = ``new` `Node(``8``); ` `        ``root.left.right.right = ``new` `Node(``9``); ` ` `  `        ``// Find and print the path ` `        ``GFG pathBetweenNodes = ``new` `GFG(root); ` `        ``Stack stack ` `            ``= pathBetweenNodes.pathBetweenNode(``7``, ``4``); ` `        ``print(stack); ` `    ``} ` `} `

Output:

```7 3 1 4
```

My Personal Notes arrow_drop_up Software Engineer, an Algorithms enthusiast

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 Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.