Update readme.md

Co-Authored-By: Antim-IWP <203163676+Antim-IWP@users.noreply.github.com>
Co-Authored-By: Shiwangi Srivastava <174641070+IamShiwangi@users.noreply.github.com>

Author: 3 people

Solution/1448. Count Good Nodes in Binary Tree/readme.md

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+Sure! Here's the full `README.md` file for **LeetCode Problem 1448. Count Good Nodes in Binary Tree**, following the exact style you requested, like the `doocs/leetcode` GitHub repo:
+
+---
+
+```markdown
+# [1448. Count Good Nodes in Binary Tree](https://leetcode.com/problems/count-good-nodes-in-binary-tree)
+
+> Difficulty: Medium  
+> Biweekly Contest 26 Q3  
+> Tags: Tree, Depth-First Search, Breadth-First Search, Binary Tree
+
+## Description
+
+Given a binary tree `root`, a node **X** in the tree is named **good** if in the path from root to **X** there are no nodes with a value **greater than** X.
+
+Return the number of **good** nodes in the binary tree.
+
+### Example 1:
+
+![example1](https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1448.Count%20Good%20Nodes%20in%20Binary%20Tree/images/test_sample_1.png)
+
+```
+Input: root = [3,1,4,3,null,1,5]
+Output: 4
+Explanation: Nodes in blue are good.
+Root Node (3) is always a good node.
+Node 4 -> (3,4) is the maximum value in the path starting from the root.
+Node 5 -> (3,4,5) is the maximum value in the path
+Node 3 -> (3,1,3) is the maximum value in the path.
+```
+
+### Example 2:
+
+![example2](https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1448.Count%20Good%20Nodes%20in%20Binary%20Tree/images/test_sample_2.png)
+
+```
+Input: root = [3,3,null,4,2]
+Output: 3
+Explanation: Node 2 -> (3, 3, 2) is not good, because "3" is higher than it.
+```
+
+### Example 3:
+
+```
+Input: root = [1]
+Output: 1
+Explanation: Root is considered as good.
+```
+
+### Constraints:
+
+- The number of nodes in the binary tree is in the range `[1, 10^5]`.
+- Each node's value is between `[-10^4, 10^4]`.
+
+## Solutions
+
+### Python3
+
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def goodNodes(self, root: TreeNode) -> int:
+        def dfs(root: TreeNode, mx: int):
+            if root is None:
+                return
+            nonlocal ans
+            if mx <= root.val:
+                ans += 1
+                mx = root.val
+            dfs(root.left, mx)
+            dfs(root.right, mx)
+
+        ans = 0
+        dfs(root, -1000000)
+        return ans
+```
+
+### Java
+
+```java
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+class Solution {
+    private int ans = 0;
+
+    public int goodNodes(TreeNode root) {
+        dfs(root, -100000);
+        return ans;
+    }
+
+    private void dfs(TreeNode root, int mx) {
+        if (root == null) {
+            return;
+        }
+        if (mx <= root.val) {
+            ++ans;
+            mx = root.val;
+        }
+        dfs(root.left, mx);
+        dfs(root.right, mx);
+    }
+}
+```
+
+### C++
+
+```cpp
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    int goodNodes(TreeNode* root) {
+        int ans = 0;
+        function<void(TreeNode*, int)> dfs = [&](TreeNode* root, int mx) {
+            if (!root) {
+                return;
+            }
+            if (mx <= root->val) {
+                ++ans;
+                mx = root->val;
+            }
+            dfs(root->left, mx);
+            dfs(root->right, mx);
+        };
+        dfs(root, -1e6);
+        return ans;
+    }
+};
+```
+
+### Go
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func goodNodes(root *TreeNode) (ans int) {
+	var dfs func(*TreeNode, int)
+	dfs = func(root *TreeNode, mx int) {
+		if root == nil {
+			return
+		}
+		if mx <= root.Val {
+			ans++
+			mx = root.Val
+		}
+		dfs(root.Left, mx)
+		dfs(root.Right, mx)
+	}
+	dfs(root, -10001)
+	return
+}
+```
+
+### TypeScript
+
+```ts
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *     val: number
+ *     left: TreeNode | null
+ *     right: TreeNode | null
+ *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.left = (left===undefined ? null : left)
+ *         this.right = (right===undefined ? null : right)
+ *     }
+ * }
+ */
+
+function goodNodes(root: TreeNode | null): number {
+    let ans = 0;
+    const dfs = (root: TreeNode | null, mx: number) => {
+        if (!root) {
+            return;
+        }
+        if (mx <= root.val) {
+            ++ans;
+            mx = root.val;
+        }
+        dfs(root.left, mx);
+        dfs(root.right, mx);
+    };
+    dfs(root, -1e6);
+    return ans;
+}
+```
+```
+
+---
+
+Let me know if you want the Chinese version (`README.md`) or to continue with more problems in this format!