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# Find the parent of a node in the given binary tree

• Difficulty Level : Medium
• Last Updated : 13 Aug, 2021

Given a tree and a node, the task is to find the parent of the given node in the tree. Print -1 if the given node is the root node.
Examples:

```Input: Node = 3
1
/   \
2     3
/ \
4   5
Output: 1

Input: Node = 1
1
/   \
2     3
/       \
4         5
/
6
Output: -1```

Approach: Write a recursive function that takes the current node and its parent as the arguments (root node is passed with -1 as its parent). If the current node is equal to the required node then print its parent and return else call the function recursively for its children and the current node as the parent.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``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;``    ``Node(``int` `data)``    ``{``        ``this``->data = data;``        ``left = right = NULL;``    ``}``};` `// Recursive function to find the``// parent of the given node``void` `findParent(``struct` `Node* node,``                ``int` `val, ``int` `parent)``{``    ``if` `(node == NULL)``        ``return``;` `    ``// If current node is the required node``    ``if` `(node->data == val) {` `        ``// Print its parent``        ``cout << parent;``    ``}``    ``else` `{` `        ``// Recursive calls for the children``        ``// of the current node``        ``// Current node is now the new parent``        ``findParent(node->left, val, node->data);``        ``findParent(node->right, val, node->data);``    ``}``}` `// Driver code``int` `main()``{``    ``struct` `Node* root = ``new` `Node(1);``    ``root->left = ``new` `Node(2);``    ``root->right = ``new` `Node(3);``    ``root->left->left = ``new` `Node(4);``    ``root->left->right = ``new` `Node(5);``    ``int` `node = 3;` `    ``findParent(root, node, -1);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `/* A binary tree node has data, pointer``to left child and a pointer``to right child */``static` `class` `Node``{``    ``int` `data;``    ``Node left, right;``    ``Node(``int` `data)``    ``{``        ``this``.data = data;``        ``left = right = ``null``;``    ``}``};` `// Recursive function to find the``// parent of the given node``static` `void` `findParent(Node node,``                       ``int` `val, ``int` `parent)``{``    ``if` `(node == ``null``)``        ``return``;` `    ``// If current node is the required node``    ``if` `(node.data == val)``    ``{` `        ``// Print its parent``        ``System.out.print(parent);``    ``}``    ``else``    ``{` `        ``// Recursive calls for the children``        ``// of the current node``        ``// Current node is now the new parent``        ``findParent(node.left, val, node.data);``        ``findParent(node.right, val, node.data);``    ``}``}` `// Driver code``public` `static` `void` `main(String []args)``{``    ``Node root = ``new` `Node(``1``);``    ``root.left = ``new` `Node(``2``);``    ``root.right = ``new` `Node(``3``);``    ``root.left.left = ``new` `Node(``4``);``    ``root.left.right = ``new` `Node(``5``);``    ``int` `node = ``3``;` `    ``findParent(root, node, -``1``);``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of``# the above approach` `''' A binary tree node has data, pointer``to left child and a pointer``to right child '''``class` `Node:``  ` `    ``def` `__init__(``self``, data):``      ` `        ``self``.data ``=` `data``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# Recursive function to find the``# parent of the given node``def` `findParent(node : Node,``               ``val : ``int``,``               ``parent : ``int``) ``-``> ``None``:``    ``if` `(node ``is` `None``):``        ``return` `    ``# If current node is``    ``# the required node``    ``if` `(node.data ``=``=` `val):``      ` `        ``# Print its parent``        ``print``(parent)``    ``else``:``      ` `        ``# Recursive calls``        ``# for the children``        ``# of the current node``        ``# Current node is now``        ``# the new parent``        ``findParent(node.left,``                   ``val, node.data)``        ``findParent(node.right,``                   ``val, node.data)` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``root ``=` `Node(``1``)``    ``root.left ``=` `Node(``2``)``    ``root.right ``=` `Node(``3``)``    ``root.left.left ``=` `Node(``4``)``    ``root.left.right ``=` `Node(``5``)``    ``node ``=` `3``    ``findParent(root, node, ``-``1``)` `# This code is contributed by sanjeev2552`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{` `/* 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``;``    ``}``};` `// Recursive function to find the``// parent of the given node``static` `void` `findParent(Node node,``                         ``int` `val, ``int` `parent)``{``    ``if` `(node == ``null``)``        ``return``;` `    ``// If current node is the required node``    ``if` `(node.data == val)``    ``{` `        ``// Print its parent``        ``Console.Write(parent);``    ``}``    ``else``    ``{` `        ``// Recursive calls for the children``        ``// of the current node``        ``// Current node is now the new parent``        ``findParent(node.left, val, node.data);``        ``findParent(node.right, val, node.data);``    ``}``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``Node root = ``new` `Node(1);``    ``root.left = ``new` `Node(2);``    ``root.right = ``new` `Node(3);``    ``root.left.left = ``new` `Node(4);``    ``root.left.right = ``new` `Node(5);``    ``int` `node = 3;` `    ``findParent(root, node, -1);``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

`1`

Time Complexity: O(N).
Auxiliary Space: O(N).

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