feat: add python and java solutions to lcof problem

添加《剑指 Offer》题解：面试题32 - I. 从上到下打印二叉树

Author: yanglbme

lcof/面试题32 - I. 从上到下打印二叉树/README.md

@@ -0,0 +1,101 @@
+# [面试题32 - I. 从上到下打印二叉树](https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-lcof/)
+
+## 题目描述
+从上到下打印出二叉树的每个节点，同一层的节点按照从左到右的顺序打印。
+
+**例如:**
+
+给定二叉树: `[3,9,20,null,null,15,7]`,
+
+```
+    3
+   / \
+  9  20
+    /  \
+   15   7
+```
+
+**返回：**
+
+```
+[3,9,20,15,7]
+```
+
+**提示：**
+
+- `节点总数 <= 1000`
+
+## 解法
+### Python3
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+from queue import Queue
+
+class Solution:
+    def levelOrder(self, root: TreeNode) -> List[int]:
+        if root is None:
+            return []
+        s = Queue()
+        res = []
+        s.put(root)
+        while not s.empty():
+            node = s.get()
+            res.append(node.val)
+            if node.left:
+                s.put(node.left)
+            if node.right:
+                s.put(node.right)
+        return res            
+```
+
+### Java
+```java
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public int[] levelOrder(TreeNode root) {
+        if (root == null) {
+            return new int[0];
+        }
+        Queue<TreeNode> q = new LinkedList<>();
+        Queue<Integer> s = new LinkedList<>();
+        q.offer(root);
+
+        while (!q.isEmpty()) {
+            TreeNode node = q.poll();
+            s.offer(node.val);
+            if (node.left != null) {
+                q.offer(node.left);
+            }
+            if (node.right != null) {
+                q.offer(node.right);
+            }
+        }
+        int[] res = new int[s.size()];
+
+        for (int i = 0, len = s.size(); i < len; ++i) {
+            res[i] = s.poll();
+        }
+        return res;
+
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题32 - I. 从上到下打印二叉树/Solution.java

@@ -0,0 +1,37 @@
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public int[] levelOrder(TreeNode root) {
+        if (root == null) {
+            return new int[0];
+        }
+        Queue<TreeNode> q = new LinkedList<>();
+        Queue<Integer> s = new LinkedList<>();
+        q.offer(root);
+
+        while (!q.isEmpty()) {
+            TreeNode node = q.poll();
+            s.offer(node.val);
+            if (node.left != null) {
+                q.offer(node.left);
+            }
+            if (node.right != null) {
+                q.offer(node.right);
+            }
+        }
+        int[] res = new int[s.size()];
+
+        for (int i = 0, len = s.size(); i < len; ++i) {
+            res[i] = s.poll();
+        }
+        return res;
+
+    }
+}

lcof/面试题32 - I. 从上到下打印二叉树/Solution.py

@@ -0,0 +1,24 @@
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+from queue import Queue
+
+class Solution:
+    def levelOrder(self, root: TreeNode) -> List[int]:
+        if root is None:
+            return []
+        s = Queue()
+        res = []
+        s.put(root)
+        while not s.empty():
+            node = s.get()
+            res.append(node.val)
+            if node.left:
+                s.put(node.left)
+            if node.right:
+                s.put(node.right)
+        return res