Leetcode 75

Solution/104. Maximum Depth of Binary Tree/104. Maximum Depth of Binary Tree.py

@@ -0,0 +1,12 @@
+# 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 maxDepth(self, root: TreeNode) -> int:
+        if root is None:
+            return 0
+        l, r = self.maxDepth(root.left), self.maxDepth(root.right)
+        return 1 + max(l, r)

Solution/104. Maximum Depth of Binary Tree/readme.md

@@ -0,0 +1,268 @@
+
+<!-- problem:start -->
+
+# [104. Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree)
+
+---
+- **comments**: true
+- **difficulty**: Easy
+- **edit_url**: https://github.com/doocs/leetcode/edit/main/solution/0100-0199/0104.Maximum%20Depth%20of%20Binary%20Tree/README_EN.md
+- **tags**:
+    - Tree
+    - Depth-First Search
+    - Breadth-First Search
+    - Binary Tree
+---
+
+
+## Description
+
+<!-- description:start -->
+
+<p>Given the <code>root</code> of a binary tree, return <em>its maximum depth</em>.</p>
+
+<p>A binary tree&#39;s <strong>maximum depth</strong>&nbsp;is the number of nodes along the longest path from the root node down to the farthest leaf node.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0104.Maximum%20Depth%20of%20Binary%20Tree/images/tmp-tree.jpg" style="width: 400px; height: 277px;" />
+<pre>
+<strong>Input:</strong> root = [3,9,20,null,null,15,7]
+<strong>Output:</strong> 3
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> root = [1,null,2]
+<strong>Output:</strong> 2
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the tree is in the range <code>[0, 10<sup>4</sup>]</code>.</li>
+	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Recursion
+
+Recursively traverse the left and right subtrees, calculate the maximum depth of the left and right subtrees, and then take the maximum value plus $1$.
+
+The time complexity is $O(n)$, where $n$ is the number of nodes in the binary tree. Each node is traversed only once in the recursion.
+
+<!-- tabs:start -->
+
+#### 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 maxDepth(self, root: TreeNode) -> int:
+        if root is None:
+            return 0
+        l, r = self.maxDepth(root.left), self.maxDepth(root.right)
+        return 1 + max(l, r)
+```
+
+#### 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 {
+    public int maxDepth(TreeNode root) {
+        if (root == null) {
+            return 0;
+        }
+        int l = maxDepth(root.left);
+        int r = maxDepth(root.right);
+        return 1 + Math.max(l, r);
+    }
+}
+```
+
+#### 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 maxDepth(TreeNode* root) {
+        if (!root) return 0;
+        int l = maxDepth(root->left), r = maxDepth(root->right);
+        return 1 + max(l, r);
+    }
+};
+```
+
+#### Go
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func maxDepth(root *TreeNode) int {
+	if root == nil {
+		return 0
+	}
+	l, r := maxDepth(root.Left), maxDepth(root.Right)
+	return 1 + max(l, r)
+}
+```
+
+#### 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 maxDepth(root: TreeNode | null): number {
+    if (root === null) {
+        return 0;
+    }
+    return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
+}
+```
+
+#### Rust
+
+```rust
+// Definition for a binary tree node.
+// #[derive(Debug, PartialEq, Eq)]
+// pub struct TreeNode {
+//   pub val: i32,
+//   pub left: Option<Rc<RefCell<TreeNode>>>,
+//   pub right: Option<Rc<RefCell<TreeNode>>>,
+// }
+//
+// impl TreeNode {
+//   #[inline]
+//   pub fn new(val: i32) -> Self {
+//     TreeNode {
+//       val,
+//       left: None,
+//       right: None
+//     }
+//   }
+// }
+use std::cell::RefCell;
+use std::rc::Rc;
+impl Solution {
+    fn dfs(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
+        if root.is_none() {
+            return 0;
+        }
+        let node = root.as_ref().unwrap().borrow();
+        1 + Self::dfs(&node.left).max(Self::dfs(&node.right))
+    }
+
+    pub fn max_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
+        Self::dfs(&root)
+    }
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {number}
+ */
+var maxDepth = function (root) {
+    if (!root) return 0;
+    const l = maxDepth(root.left);
+    const r = maxDepth(root.right);
+    return 1 + Math.max(l, r);
+};
+```
+
+#### C
+
+```c
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     struct TreeNode *left;
+ *     struct TreeNode *right;
+ * };
+ */
+
+#define max(a, b) (((a) > (b)) ? (a) : (b))
+
+int maxDepth(struct TreeNode* root) {
+    if (!root) {
+        return 0;
+    }
+    int left = maxDepth(root->left);
+    int right = maxDepth(root->right);
+    return 1 + max(left, right);
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

Solution/1372. Longest ZigZag Path in a Binary Tree/1372. Longest ZigZag Path in a Binary Tree.py

@@ -0,0 +1,19 @@
+# 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 longestZigZag(self, root: TreeNode) -> int:
+        def dfs(root, l, r):
+            if root is None:
+                return
+            nonlocal ans
+            ans = max(ans, l, r)
+            dfs(root.left, r + 1, 0)
+            dfs(root.right, 0, l + 1)
+
+        ans = 0
+        dfs(root, 0, 0)
+        return ans