Added solution for leetcode problem 0111

dsa-solutions/lc-solutions/0100-0199/0111-minimum-depth.md

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+---
+id:  minimum-depth-of-binary-tree
+title:  Minimum Depth of Binary Tree
+sidebar_label: 0111-Minimum Depth Of Binary tree
+tags:
+  - Java
+  - Python
+  - C++
+description: "Find the minimum depth of a binary tree, which is the number of nodes along the shortest path from the root node down to the nearest leaf node."
+---
+
+## Problem Description
+
+Given a binary tree, find its minimum depth.
+
+The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
+
+### Examples
+
+**Example 1:**
+
+![LeetCode Problem - Binary Tree](https://assets.leetcode.com/uploads/2020/10/12/ex_depth.jpg)
+```
+Input: root = [3,9,20,null,null,15,7]
+Output: 2
+```
+
+
+### Constraints
+
+- The number of nodes in the tree is in the range [0, 105].
+- -100 <= Node.val <= 100
+
+---
+
+## Solution for Binary Tree Problem
+
+### Intuition And Approach
+
+To find the minimum depth of a binary tree, we can perform a depth-first search (DFS) or a breadth-first search (BFS). Here, we will use BFS for simplicity.
+
+Breadth-First Search (BFS): Traverse the tree level by level, stopping as soon as we encounter a leaf node. The depth of the first leaf node encountered will be the minimum depth of the tree.
+
+<Tabs>
+ <tabItem value="Recursive" label="Recursive">
+
+
+#### Code in Different Languages
+
+<Tabs>
+  <TabItem value="Java" label="Java" default>
+  <SolutionAuthor name="@Vipullakum007"/>
+   ```java
+   class Solution {
+    public int minDepth(TreeNode root) {
+        if (root == null) return 0;
+        Queue<TreeNode> queue = new LinkedList<>();
+        queue.offer(root);
+        int depth = 1;
+        while (!queue.isEmpty()) {
+            int levelSize = queue.size();
+            for (int i = 0; i < levelSize; i++) {
+                TreeNode node = queue.poll();
+                if (node.left == null && node.right == null) return depth;
+                if (node.left != null) queue.offer(node.left);
+                if (node.right != null) queue.offer(node.right);
+            }
+            depth++;
+        }
+        return depth;
+    }
+}
+    ```
+
+  </TabItem>
+  <TabItem value="Python" label="Python">
+  <SolutionAuthor name="@Vipullakum007"/>
+   ```python
+    class Solution:
+    def minDepth(self, root):
+        if not root:
+            return 0
+        queue = collections.deque([(root, 1)])
+        while queue:
+            node, depth = queue.popleft()
+            if not node.left and not node.right:
+                return depth
+            if node.left:
+                queue.append((node.left, depth + 1))
+            if node.right:
+                queue.append((node.right, depth + 1))
+    ```
+
+  </TabItem>
+  <TabItem value="C++" label="C++">
+  <SolutionAuthor name="@Vipullakum007"/>
+   ```cpp
+    class Solution {
+public:
+    int minDepth(TreeNode* root) {
+        if (!root) return 0;
+        queue<TreeNode*> q;
+        q.push(root);
+        int depth = 1;
+        while (!q.empty()) {
+            int levelSize = q.size();
+            for (int i = 0; i < levelSize; ++i) {
+                TreeNode* node = q.front();
+                q.pop();
+                if (!node->left && !node->right) return depth; // Leaf node found
+                if (node->left) q.push(node->left);
+                if (node->right) q.push(node->right);
+            }
+            ++depth;
+        }
+        return depth;
+    }
+};
+    ```
+
+  </TabItem>
+</Tabs>
+
+#### Complexity Analysis
+
+- Time Complexity: $O(n)$ where n is the number of nodes in the binary tree.
+- Space Complexity: $O(h)$ where h is the height of the binary tree.
+
+</tabItem>
+</Tabs>
+
+
+---