feat: solve No.543

Author: BaffinLee

501-600/543. Diameter of Binary Tree.md

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+# 543. Diameter of Binary Tree
+
+- Difficulty: Easy.
+- Related Topics: Tree, Depth-First Search, Binary Tree.
+- Similar Questions: Diameter of N-Ary Tree, Longest Path With Different Adjacent Characters.
+
+## Problem
+
+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.
+
+ 
+Example 1:
+
+![](https://assets.leetcode.com/uploads/2021/03/06/diamtree.jpg)
+
+```
+Input: root = [1,2,3,4,5]
+Output: 3
+Explanation: 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, 104]`.
+	
+- `-100 <= Node.val <= 100`
+
+
+
+## Solution
+
+```javascript
+/**
+ * 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 diameterOfBinaryTree = function(root) {
+    return dfs(root)[0];
+};
+
+var dfs = function(node) {
+    if (!node) return [0, 0];
+    var left = dfs(node.left);
+    var right = dfs(node.right);
+    var child = Math.max(
+        node.left ? left[1] + 1 : 0,
+        node.right ? right[1] + 1 : 0,
+    );
+    var branch = (node.left ? left[1] + 1 : 0) +
+        (node.right ? right[1] + 1 : 0);
+    return [
+        Math.max(
+            left[0],
+            right[0],
+            branch,
+        ),
+        child,
+    ];
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(1).