modify code

Author: algorithmzuo

src/class036/Code01_LevelOrderTraversal.java

@@ -17,7 +17,7 @@ public static class TreeNode {
 		public TreeNode right;
 	}
 
-	// 提交时把方法名改为levelOrder，此方法为普通bfs，不推荐
+	// 提交时把方法名改为levelOrder，此方法为普通bfs，此题不推荐
 	public static List<List<Integer>> levelOrder1(TreeNode root) {
 		List<List<Integer>> ans = new ArrayList<>();
 		if (root != null) {
@@ -52,13 +52,13 @@ public static List<List<Integer>> levelOrder1(TreeNode root) {
 
 	public static int l, r;
 
-	// 提交时把方法名改为levelOrder，此方法为每次处理一层的优化bfs，推荐
+	// 提交时把方法名改为levelOrder，此方法为每次处理一层的优化bfs，此题推荐
 	public static List<List<Integer>> levelOrder2(TreeNode root) {
 		List<List<Integer>> ans = new ArrayList<>();
 		if (root != null) {
 			l = r = 0;
 			queue[r++] = root;
-			while (l < r) {
+			while (l < r) { // 队列里还有东西
 				int size = r - l;
 				ArrayList<Integer> list = new ArrayList<Integer>();
 				for (int i = 0; i < size; i++) {

src/class036/Code02_ZigzagLevelOrderTraversal.java

@@ -28,10 +28,16 @@ public static List<List<Integer>> zigzagLevelOrder(TreeNode root) {
 		if (root != null) {
 			l = r = 0;
 			queue[r++] = root;
-			boolean reverse = false;
+			// false 代表从左往右
+			// true 代表从右往左
+			boolean reverse = false; 
 			while (l < r) {
 				int size = r - l;
 				ArrayList<Integer> list = new ArrayList<Integer>();
+				// reverse == false, 左 -> 右， l....r-1, 收集size个
+				// reverse == true,  右 -> 左， r-1....l, 收集size个
+				// 左 -> 右, i = i + 1
+				// 右 -> 左, i = i - 1
 				for (int i = reverse ? r - 1 : l, j = reverse ? -1 : 1, k = 0; k < size; i += j, k++) {
 					TreeNode cur = queue[i];
 					list.add(cur.val);

src/class036/Code04_DepthOfBinaryTree.java

@@ -18,9 +18,11 @@ public static int maxDepth(TreeNode root) {
 	// 测试链接 : https://leetcode.cn/problems/minimum-depth-of-binary-tree/
 	public int minDepth(TreeNode root) {
 		if (root == null) {
+			// 当前的树是空树
 			return 0;
 		}
 		if (root.left == null && root.right == null) {
+			// 当前root是叶节点
 			return 1;
 		}
 		int ldeep = Integer.MAX_VALUE;

src/class036/Code05_PreorderSerializeAndDeserialize.java

@@ -32,36 +32,37 @@ public class Codec {
 
 		public String serialize(TreeNode root) {
 			StringBuilder builder = new StringBuilder();
-			serialize(root, builder);
+			f(root, builder);
 			return builder.toString();
 		}
 
-		void serialize(TreeNode root, StringBuilder builder) {
+		void f(TreeNode root, StringBuilder builder) {
 			if (root == null) {
 				builder.append("#,");
 			} else {
 				builder.append(root.val + ",");
-				serialize(root.left, builder);
-				serialize(root.right, builder);
+				f(root.left, builder);
+				f(root.right, builder);
 			}
 		}
 
 		public TreeNode deserialize(String data) {
 			String[] vals = data.split(",");
 			cnt = 0;
-			return deserialize(vals);
+			return g(vals);
 		}
 
+		// 当前数组消费到哪了
 		public static int cnt;
 
-		TreeNode deserialize(String[] vals) {
+		TreeNode g(String[] vals) {
 			String cur = vals[cnt++];
 			if (cur.equals("#")) {
 				return null;
 			} else {
 				TreeNode head = new TreeNode(Integer.valueOf(cur));
-				head.left = deserialize(vals);
-				head.right = deserialize(vals);
+				head.left = g(vals);
+				head.right = g(vals);
 				return head;
 			}
 		}

src/class036/Code07_PreorderInorderBuildBinaryTree.java

@@ -37,9 +37,12 @@ public static TreeNode f(int[] pre, int l1, int r1, int[] in, int l2, int r2, Ha
 		if (l1 == r1) {
 			return head;
 		}
-		int find = map.get(pre[l1]);
-		head.left = f(pre, l1 + 1, l1 + find - l2, in, l2, find - 1, map);
-		head.right = f(pre, l1 + find - l2 + 1, r1, in, find + 1, r2, map);
+		int k = map.get(pre[l1]);
+		// pre : l1(........)[.......r1]
+		// in  : (l2......)k[........r2]
+		// (...)是左树对应，[...]是右树的对应
+		head.left = f(pre, l1 + 1, l1 + k - l2, in, l2, k - 1, map);
+		head.right = f(pre, l1 + k - l2 + 1, r1, in, k + 1, r2, map);
 		return head;
 	}
 

src/class036/Code09_CountCompleteTreeNodes.java

@@ -19,17 +19,25 @@ public static int countNodes(TreeNode head) {
 		return f(head, 1, mostLeft(head, 1));
 	}
 
-	public static int f(TreeNode cur, int Level, int h) {
-		if (Level == h) {
+	// cur : 当前来到的节点
+	// level :  当前来到的节点在第几层
+	// h : 整棵树的高度，不是cur这棵子树的高度
+	// 求 : cur这棵子树上有多少节点
+	public static int f(TreeNode cur, int level, int h) {
+		if (level == h) {
 			return 1;
 		}
-		if (mostLeft(cur.right, Level + 1) == h) {
-			return (1 << (h - Level)) + f(cur.right, Level + 1, h);
+		if (mostLeft(cur.right, level + 1) == h) {
+			// cur右树上的最左节点，扎到了最深层
+			return (1 << (h - level)) + f(cur.right, level + 1, h);
 		} else {
-			return (1 << (h - Level - 1)) + f(cur.left, Level + 1, h);
+			// cur右树上的最左节点，没扎到最深层
+			return (1 << (h - level - 1)) + f(cur.left, level + 1, h);
 		}
 	}
 
+	// 当前节点是cur，并且它在level层
+	// 返回从cur开始不停往左，能扎到几层
 	public static int mostLeft(TreeNode cur, int level) {
 		while (cur != null) {
 			level++;