Added leetcode solution 543: Issue no #1413

dsa-solutions/lc-solutions/0500-0599/0543-Diameter-of-Binary-Tree.md

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+---
+id: diameter-of-a-binary-tree
+title:   Diameter of Binary tree
+sidebar_label: 0543. Diameter of Binary tree
+
+tags:
+- Binary Tree
+- Diameter
+
+description: "This is a solution to the Diameter of Binary tree problem on LeetCode."
+---
+
+## Problem Description
+Given the `root` of a binary tree, return *the length of the diameter of the tree*.
+The **diameter** of a binary tree is the **length** of the longest path between any two nodes in a tree. This path may or may not pass through the `root`.
+The **length** of a path between two nodes is represented by the number of edges between them.
+### Examples
+
+**Example 1:** 
+
+![Example1](0543-Ex1.png)
+```
+Input: root = [1,2,3,4,5]
+Output: 3
+Explanataion: 3 is the length of the path [4,2,1,3] or [5,2,1,3].
+```
+
+**Example 2:**
+
+```
+Input: root = [1,2]
+Output: 1
+```
+
+
+### Constraints
+- The number of nodes in the tree is in the range `[1, 10^4]`.
+- `-100 <= Node.val <= 100`
+
+
+
+## Solution for Diameter of Binary tree
+### Approach 
+#### Brute Force 
+- **Define a Helper Function**: 
+    - Create a function to calculate the height of a tree.
+- **Calculate Diameter**: 
+    - For each node, calculate the diameter passing through that node.
+    - The diameter passing through a node is the sum of the heights of its left and right subtrees.
+    - Compare this diameter with the global maximum diameter.
+- **Traverse Tree**: 
+    - Perform a traversal of the tree (e.g., in-order, pre-order) to compute the diameter for each node.
+
+**Implementation:**
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+
+struct TreeNode {
+    int val;
+    TreeNode *left;
+    TreeNode *right;
+    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+};
+
+int height(TreeNode* node) {
+    if (node == NULL) {
+        return 0;
+    }
+    return 1 + max(height(node->left), height(node->right));
+}
+
+int diameterOfBinaryTree(TreeNode* root) {
+    if (root == NULL) {
+        return 0;
+    }
+
+    // Get the height of left and right sub-trees
+    int left_height = height(root->left);
+    int right_height = height(root->right);
+
+    // Get the diameter of left and right sub-trees
+    int left_diameter = diameterOfBinaryTree(root->left);
+    int right_diameter = diameterOfBinaryTree(root->right);
+
+    // Calculate diameter passing through the root
+    return max(left_height + right_height, max(left_diameter, right_diameter));
+}
+
+int main() {
+    TreeNode* root = new TreeNode(1);
+    root->left = new TreeNode(2);
+    root->right = new TreeNode(3);
+    root->left->left = new TreeNode(4);
+    root->left->right = new TreeNode(5);
+
+    cout << "Diameter of the binary tree is: " << diameterOfBinaryTree(root) << endl;
+
+    return 0;
+}
+```
+
+**Complexity:**
+- Time Complexity: `O(n^2)` because for each node, we are calculating the height which takes O(n) time.
+- Space Complexity: `O(n)` due to the recursion stack.
+
+**Corner Cases:**
+- Empty tree: Should return `0`.
+- Single node tree: Should return `0` as there are no edges.
+
+#### Optimized Approach 
+- **Define a Helper Function:**: 
+    - Create a function to calculate both the height and the diameter of the tree simultaneously.
+- **Single Traversal**: 
+    - Traverse the tree once, calculating the height and updating the maximum diameter at each node.
+
+**Implementation:**
+
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+
+struct TreeNode {
+    int val;
+    TreeNode *left;
+    TreeNode *right;
+    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+};
+
+class Solution {
+public:
+    int diameter;
+
+    int height(TreeNode* node) {
+        if (node == NULL) {
+            return 0;
+        }
+
+        int left_height = height(node->left);
+        int right_height = height(node->right);
+
+        // Update the diameter if left_height + right_height is larger
+        diameter = max(diameter, left_height + right_height);
+
+        // Height of the current node is max of heights of left and right subtrees plus 1
+        return 1 + max(left_height, right_height);
+    }
+
+    int diameterOfBinaryTree(TreeNode* root) {
+        diameter = 0;
+        height(root);
+        return diameter;
+    }
+};
+
+int main() {
+    TreeNode* root = new TreeNode(1);
+    root->left = new TreeNode(2);
+    root->right = new TreeNode(3);
+    root->left->left = new TreeNode(4);
+    root->left->right = new TreeNode(5);
+
+    Solution solution;
+    cout << "Diameter of the binary tree is: " << solution.diameterOfBinaryTree(root) << endl;
+
+    return 0;
+}
+```
+
+**Complexity:**
+- Time Complexity: `O(n)`since we are traversing each node only once.
+- Space Complexity: `O(n)`due to the recursion stack.
+
+**Corner Cases:**
+- Empty tree: Should return `0`.
+- Single node tree: Should return `0` as there are no edges.
+
+
+  ## Code in Different Languages
+
+ <Tabs>
+
+  <TabItem value="JavaScript" label="JavaScript">
+  <SolutionAuthor name="@vansh-codes" />
+
+   ```javascript
+    var diameterOfBinaryTree = function(root) {
+         let diameter = 0;
+
+        function height(node) {
+            if (node === null) return 0;
+            let leftHeight = height(node.left);
+            let rightHeight = height(node.right);
+
+            // Update the diameter
+            diameter = Math.max(diameter, leftHeight + rightHeight);
+
+            return 1 + Math.max(leftHeight, rightHeight);
+        }
+
+        height(root);
+        return diameter;
+    };
+    ```
+
+  </TabItem>
+
+  <TabItem value="TypeScript" label="TypeScript">
+  <SolutionAuthor name="@vansh-codes" /> 
+
+   ```typescript
+    function diameterOfBinaryTree(root: TreeNode | null): number {
+        let diameter = 0;
+
+        function height(node: TreeNode | null): number {
+            if (node === null) return 0;
+            let leftHeight = height(node.left);
+            let rightHeight = height(node.right);
+
+            // Update the diameter
+            diameter = Math.max(diameter, leftHeight + rightHeight);
+
+            return 1 + Math.max(leftHeight, rightHeight);
+        }
+
+        height(root);
+        return diameter;
+    }
+    ```
+
+  </TabItem>
+
+  <TabItem value="Python" label="Python">
+  <SolutionAuthor name="@vansh-codes" />
+
+   ```python
+    class Solution(object):
+    def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
+        self.diameter = 0
+
+        def height(node: TreeNode) -> int:
+            if not node:
+                return 0
+            left_height = height(node.left)
+            right_height = height(node.right)
+
+            # Update the diameter
+            self.diameter = max(self.diameter, left_height + right_height)
+
+            return 1 + max(left_height, right_height)
+
+        height(root)
+        return self.diameter
+    ```
+
+  </TabItem>
+
+  <TabItem value="Java" label="Java">
+  <SolutionAuthor name="@vansh-codes" />
+
+   ```java
+   import java.util.Arrays;
+
+    class Solution {
+        private int diameter;
+        public int diameterOfBinaryTree(TreeNode root) {
+            diameter = 0;
+
+            height(root);
+            return diameter;
+        }
+
+        private int height(TreeNode node) {
+            if (node == null) {
+                return 0;
+            }
+
+            int leftHeight = height(node.left);
+            int rightHeight = height(node.right);
+
+            // Update the diameter
+            diameter = Math.max(diameter, leftHeight + rightHeight);
+
+            return 1 + Math.max(leftHeight, rightHeight);
+        }
+    };
+    ```
+
+  </TabItem>
+
+ <TabItem value="C++" label="C++">
+ <SolutionAuthor name="@vansh-codes" />
+
+   ```cpp
+   class Solution {
+    public:
+        int diameter;
+
+        int diameterOfBinaryTree(TreeNode* root) {
+            diameter = 0;
+            height(root);
+            return diameter;
+        }
+    private:
+        int height(TreeNode* node) {
+            if (node == nullptr) {
+                return 0;
+            }
+
+            int leftHeight = height(node->left);
+            int rightHeight = height(node->right);
+
+            // Update the diameter
+            diameter = max(diameter, leftHeight + rightHeight);
+
+            return 1 + max(leftHeight, rightHeight);
+        }
+    };
+    ```
+
+ </TabItem>
+ </Tabs>
+
+## References
+
+- **LeetCode Problem**: [Diameter of Binary tree](https://leetcode.com/problems/diameter-of-binary-tree/description/)
+
+- **Solution Link**: [LeetCode Solution](https://leetcode.com/problems/diameter-of-binary-tree/solutions/)