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# Smallest Subtree with all the Deepest Nodes

• Difficulty Level : Basic
• Last Updated : 21 Jun, 2021

Given a Binary Tree, the task is to find the smallest subtree that contains all the deepest nodes of the given Binary Tree and return the root of that subtree.
Note: Depth of each node is defined as the length of the path from the root to the given node.

Examples:

Input:

```            1
/   \
2     3
/  \   /  \
4    5  6   7
/  \
8    9```

Output: 7

Input:

```               1
/   \
2     3```

Output: 1

Approach: Follow the steps below to solve the problem:

• Traverse the Binary Tree recursively using DFS .
• For every node, find the depth of its left and right subtrees.
• If depth of the left subtree > depth of the right subtree: Traverse the left subtree.
• If depth of the right subtree > depth of the left subtree: Traverse the right subtree.
• Otherwise, return the current node.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach``#include ``using` `namespace` `std;` `// Structure of a Node``struct` `TreeNode {` `    ``int` `val;``    ``TreeNode* left;``    ``TreeNode* right;` `    ``TreeNode(``int` `data)``    ``{``        ``this``->val = data;``        ``left = NULL;``        ``right = NULL;``    ``}``};` `// Function to return depth of``// the Tree from root``int` `find_ht(TreeNode* root)``{``    ``if` `(!root)``        ``return` `0;` `    ``// If current node is a leaf node``    ``if` `(root->left == NULL``        ``&& root->right == NULL)``        ``return` `1;` `    ``return` `max(find_ht(root->left),``               ``find_ht(root->right))``           ``+ 1;``}` `// Function to find the root of the smallest``// subtree consisting of all deepest nodes``void` `find_node(TreeNode* root, TreeNode*& req_node)``{``    ``if` `(!root)``        ``return``;` `    ``// Stores height of left subtree``    ``int` `left_ht = find_ht(root->left);` `    ``// Stores height of right subtree``    ``int` `right_ht = find_ht(root->right);` `    ``// If height of left subtree exceeds``    ``// that of the right subtree``    ``if` `(left_ht > right_ht) {` `        ``// Traverse left subtree``        ``find_node(root->left, req_node);``    ``}` `    ``// If height of right subtree exceeds``    ``// that of the left subtree``    ``else` `if` `(right_ht > left_ht) {``        ``find_node(root->right, req_node);``    ``}` `    ``// Otherwise``    ``else` `{` `        ``// Return current node``        ``req_node = root;``        ``return``;``    ``}``}` `// Driver Code``int` `main()``{``    ``struct` `TreeNode* root``        ``= ``new` `TreeNode(1);``    ``root->left = ``new` `TreeNode(2);``    ``root->right = ``new` `TreeNode(3);` `    ``TreeNode* req_node = NULL;` `    ``find_node(root, req_node);` `    ``cout << req_node->val;` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;``class` `GFG``{` `// Structure of a Node``static` `class` `TreeNode``{``    ``int` `val;``    ``TreeNode left;``    ``TreeNode right;``    ``TreeNode(``int` `data)``    ``{``        ``this``.val = data;``        ``left = ``null``;``        ``right = ``null``;``    ``}``};` `// Function to return depth of``// the Tree from root``static` `int` `find_ht(TreeNode root)``{``    ``if` `(root == ``null``)``        ``return` `0``;` `    ``// If current node is a leaf node``    ``if` `(root.left == ``null``        ``&& root.right == ``null``)``        ``return` `1``;``    ``return` `Math.max(find_ht(root.left),``               ``find_ht(root.right))``           ``+ ``1``;``}` `// Function to find the root of the smallest``// subtree consisting of all deepest nodes``static` `TreeNode find_node(TreeNode root, TreeNode req_node)``{``    ``if` `(root == ``null``)``        ``return` `req_node;` `    ``// Stores height of left subtree``    ``int` `left_ht = find_ht(root.left);` `    ``// Stores height of right subtree``    ``int` `right_ht = find_ht(root.right);` `    ``// If height of left subtree exceeds``    ``// that of the right subtree``    ``if` `(left_ht > right_ht)``    ``{` `        ``// Traverse left subtree``        ``req_node = find_node(root.left, req_node);``    ``}` `    ``// If height of right subtree exceeds``    ``// that of the left subtree``    ``else` `if` `(right_ht > left_ht)``    ``{``        ``req_node = find_node(root.right, req_node);``    ``}` `    ``// Otherwise``    ``else``    ``{` `        ``// Return current node``        ``req_node = root;``        ``return` `req_node;``    ``}``    ``return` `req_node;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``TreeNode root = ``new` `TreeNode(``1``);``    ``root.left = ``new` `TreeNode(``2``);``    ``root.right = ``new` `TreeNode(``3``);``    ``TreeNode req_node = ``null``;``    ``req_node = find_node(root, req_node);``    ``System.out.print(req_node.val);``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to implement``# the above approach` `# Structure of a Node``class` `TreeNode:``    ``def` `__init__(``self``, x):``        ``self``.val ``=` `x``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# Function to return depth of``# the Tree from root``def` `find_ht(root):``    ``if` `(``not` `root):``        ``return` `0` `    ``# If current node is a leaf node``    ``if` `(root.left ``=``=` `None` `and` `root.right ``=``=` `None``):``        ``return` `1` `    ``return` `max``(find_ht(root.left), find_ht(root.right)) ``+` `1` `# Function to find the root of the smallest``# subtree consisting of all deepest nodes``def` `find_node(root):``    ``global` `req_node` `    ``if` `(``not` `root):``        ``return` `    ``# Stores height of left subtree``    ``left_ht ``=` `find_ht(root.left)` `    ``# Stores height of right subtree``    ``right_ht ``=` `find_ht(root.right)` `    ``# If height of left subtree exceeds``    ``# that of the right subtree``    ``if` `(left_ht > right_ht):` `        ``# Traverse left subtree``        ``find_node(root.left)` `    ``# If height of right subtree exceeds``    ``# that of the left subtree``    ``elif` `(right_ht > left_ht):``        ``find_node(root.right)` `    ``# Otherwise``    ``else``:` `        ``# Return current node``        ``req_node ``=` `root``        ``return` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:` `    ``root ``=` `TreeNode(``1``)``    ``root.left ``=` `TreeNode(``2``)``    ``root.right ``=` `TreeNode(``3``)``    ``req_node ``=` `None``    ``find_node(root)``    ``print``(req_node.val)` `    ``# This code is contributed by mohit kumar 29`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG``{` `// Structure of a Node``public` `class` `TreeNode``{``    ``public` `int` `val;``    ``public` `TreeNode left;``    ``public` `TreeNode right;``    ``public` `TreeNode(``int` `data)``    ``{``        ``this``.val = data;``        ``left = ``null``;``        ``right = ``null``;``    ``}``};` `// Function to return depth of``// the Tree from root``static` `int` `find_ht(TreeNode root)``{``    ``if` `(root == ``null``)``        ``return` `0;` `    ``// If current node is a leaf node``    ``if` `(root.left == ``null``        ``&& root.right == ``null``)``        ``return` `1;``    ``return` `Math.Max(find_ht(root.left),``               ``find_ht(root.right))``           ``+ 1;``}` `// Function to find the root of the smallest``// subtree consisting of all deepest nodes``static` `TreeNode find_node(TreeNode root, TreeNode req_node)``{``    ``if` `(root == ``null``)``        ``return` `req_node;` `    ``// Stores height of left subtree``    ``int` `left_ht = find_ht(root.left);` `    ``// Stores height of right subtree``    ``int` `right_ht = find_ht(root.right);` `    ``// If height of left subtree exceeds``    ``// that of the right subtree``    ``if` `(left_ht > right_ht)``    ``{` `        ``// Traverse left subtree``        ``req_node = find_node(root.left, req_node);``    ``}` `    ``// If height of right subtree exceeds``    ``// that of the left subtree``    ``else` `if` `(right_ht > left_ht)``    ``{``        ``req_node = find_node(root.right, req_node);``    ``}` `    ``// Otherwise``    ``else``    ``{` `        ``// Return current node``        ``req_node = root;``        ``return` `req_node;``    ``}``    ``return` `req_node;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``TreeNode root = ``new` `TreeNode(1);``    ``root.left = ``new` `TreeNode(2);``    ``root.right = ``new` `TreeNode(3);``    ``TreeNode req_node = ``null``;``    ``req_node = find_node(root, req_node);``    ``Console.Write(req_node.val);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:
`1`

Time Complexity: O(NlogN)
The worst-case complexity occurs for skewed Binary Tree, whose traversal of either left or right subtree requires O(N) complexity and calculating height of subtrees requires O(logN) computational complexity.
Auxiliary Space: O(1)

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