update

src/main/java/leetcode/easy/LeetCode_0111_MinimumDepthOfBinaryTree.java

 package leetcode.easy;
 
-//给定一个二叉树，找出其最小深度。
+// 笔记：
+// 给定一个二叉树，找出其最小深度。
 //
 //        最小深度是从根节点到最近叶子节点的最短路径上的节点数量。
 //
@@ -10,68 +11,68 @@ package leetcode.easy;
 // Morris遍历
 public class LeetCode_0111_MinimumDepthOfBinaryTree {
 
-	public class TreeNode {
-		TreeNode left;
-		TreeNode right;
-	}
+  public class TreeNode {
+    TreeNode left;
+    TreeNode right;
+  }
 
-	// 二叉树的递归套路
-	public static int minDepth(TreeNode head) {
-		if (head == null) {
-			return 0;
-		}
-		if (head.left == null) {
-			return minDepth(head.right) + 1;
-		}
-		if (head.right == null) {
-			return minDepth(head.left) + 1;
-		}
-		return Math.min(minDepth(head.left), minDepth(head.right)) + 1;
-	}
+  // 二叉树的递归套路
+  public static int minDepth(TreeNode head) {
+    if (head == null) {
+      return 0;
+    }
+    if (head.left == null) {
+      return minDepth(head.right) + 1;
+    }
+    if (head.right == null) {
+      return minDepth(head.left) + 1;
+    }
+    return Math.min(minDepth(head.left), minDepth(head.right)) + 1;
+  }
 
-	// morris遍历方式
-	public static int minDepth2(TreeNode head) {
-		if (head == null) {
-			return 0;
-		}
-		TreeNode cur = head;
-		TreeNode mostRight;
-		int curHeight = 0;
-		int min = Integer.MAX_VALUE;
-		while (cur != null) {
-			mostRight = cur.left;
-			if (mostRight != null) {
-				int dulplicate = 1;
-				while (mostRight.right != null && mostRight.right != cur) {
-					dulplicate++;
-					mostRight = mostRight.right;
-				}
-				if (mostRight.right == null) {
-					curHeight++;
-					mostRight.right = cur;
-					cur = cur.left;
-					continue;
-				} else {
-					if (mostRight.left == null) {
-						min = Math.min(min, curHeight);
-					}
-					curHeight -= dulplicate;
-					mostRight.right = null;
-				}
-			} else {
-				curHeight++;
-			}
-			cur = cur.right;
-		}
-		int rightMostHeight = 1;
-		TreeNode c = head;
-		while (c.right != null) {
-			rightMostHeight++;
-			c = c.right;
-		}
-		if (c.left == null) {
-			min = Math.min(min, rightMostHeight);
-		}
-		return min;
-	}
+  // morris遍历方式
+  public static int minDepth2(TreeNode head) {
+    if (head == null) {
+      return 0;
+    }
+    TreeNode cur = head;
+    TreeNode mostRight;
+    int curHeight = 0;
+    int min = Integer.MAX_VALUE;
+    while (cur != null) {
+      mostRight = cur.left;
+      if (mostRight != null) {
+        int dulplicate = 1;
+        while (mostRight.right != null && mostRight.right != cur) {
+          dulplicate++;
+          mostRight = mostRight.right;
+        }
+        if (mostRight.right == null) {
+          curHeight++;
+          mostRight.right = cur;
+          cur = cur.left;
+          continue;
+        } else {
+          if (mostRight.left == null) {
+            min = Math.min(min, curHeight);
+          }
+          curHeight -= dulplicate;
+          mostRight.right = null;
+        }
+      } else {
+        curHeight++;
+      }
+      cur = cur.right;
+    }
+    int rightMostHeight = 1;
+    TreeNode c = head;
+    while (c.right != null) {
+      rightMostHeight++;
+      c = c.right;
+    }
+    if (c.left == null) {
+      min = Math.min(min, rightMostHeight);
+    }
+    return min;
+  }
 }