refactor 543

fishercoder1534 · fishercoder1534 · commit 7095cdd275eb · 2020-07-16T17:13:39.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_543.java b/src/main/java/com/fishercoder/solutions/_543.java
@@ -2,28 +2,11 @@
 
 import com.fishercoder.common.classes.TreeNode;
 
-/**
- * 543. Diameter of Binary Tree
- *
- * Given a binary tree, you need to compute 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.
-
- Example:
- Given a binary tree
-     1
-    / \
-   2   3
-  / \
- 4   5
- Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].
-
- Note: The length of path between two nodes is represented by the number of edges between them.
- */
 public class _543 {
 
     public static class Solution1 {
-        /**This is a very great problem for practicing recursion:
+        /**
+         * This is a very great problem for practicing recursion:
          * 1. What dfs() returns is the max height it should pick from either its left or right subtree, that's
          * what the int return type stands for;
          * 2. And during the recursion, we can keep updating the global variable: "diameter";
@@ -32,7 +15,7 @@ public static class Solution1 {
          * int left = dfs(root.left);
          * instead of dfs(root.left) + 1;
          * we'll only plus one at the end
-         * */
+         */
         int diameter = 0;
 
         public int diameterOfBinaryTree(TreeNode root) {
