feat: add python and java solutions to lcof problem

添加《剑指 Offer》题解：面试题32 - III. 从上到下打印二叉树 III

Author: yanglbme

lcof/面试题32 - III. 从上到下打印二叉树 III/README.md

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+# [面试题32 - III. 从上到下打印二叉树 III](https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-iii-lcof/)
+
+## 题目描述
+请实现一个函数按照之字形顺序打印二叉树，即第一行按照从左到右的顺序打印，第二层按照从右到左的顺序打印，第三行再按照从左到右的顺序打印，其他行以此类推。
+
+**例如:**
+
+给定二叉树: `[3,9,20,null,null,15,7]`,
+
+```
+    3
+   / \
+  9  20
+    /  \
+   15   7
+```
+
+返回其层次遍历结果：
+
+```
+[
+  [3],
+  [20,9],
+  [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[List[int]]:
+        if root is None:
+            return []
+        q = Queue()
+        q.put(root)
+        cnt = 1
+        res = []
+        level = 0
+        while not q.empty():
+            level += 1
+            t = []
+            num = 0
+            for _ in range(cnt):
+                node = q.get()
+                t.append(node.val)
+                if node.left:
+                    q.put(node.left)
+                    num += 1
+                if node.right:
+                    q.put(node.right)
+                    num += 1
+            if (level & 1) == 0:
+                t.reverse()
+            res.append(t)
+            cnt = num
+        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 List<List<Integer>> levelOrder(TreeNode root) {
+        if (root == null) {
+            return new ArrayList<>();
+        }
+        Queue<TreeNode> q = new LinkedList<>();
+        q.offer(root);
+        int cnt = 1;
+        int level = 0;
+        List<List<Integer>> res = new ArrayList<>();
+        while (!q.isEmpty()) {
+            ++level;
+            int num = 0;
+            List<Integer> t = new ArrayList<>();
+            while (cnt-- > 0) {
+                TreeNode node = q.poll();
+                t.add(node.val);
+                if (node.left != null) {
+                    q.offer(node.left);
+                    ++num;
+                }
+                if (node.right != null) {
+                    q.offer(node.right);
+                    ++num;
+                }
+            }
+            if ((level & 1) == 0) {
+                Collections.reverse(t);
+            }
+            res.add(t);
+            cnt = num;
+        }
+        return res;
+    }
+}
+```
+
+### ...
+```
+
+```

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

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+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public List<List<Integer>> levelOrder(TreeNode root) {
+        if (root == null) {
+            return new ArrayList<>();
+        }
+        Queue<TreeNode> q = new LinkedList<>();
+        q.offer(root);
+        int cnt = 1;
+        int level = 0;
+        List<List<Integer>> res = new ArrayList<>();
+        while (!q.isEmpty()) {
+            ++level;
+            int num = 0;
+            List<Integer> t = new ArrayList<>();
+            while (cnt-- > 0) {
+                TreeNode node = q.poll();
+                t.add(node.val);
+                if (node.left != null) {
+                    q.offer(node.left);
+                    ++num;
+                }
+                if (node.right != null) {
+                    q.offer(node.right);
+                    ++num;
+                }
+            }
+            if ((level & 1) == 0) {
+                Collections.reverse(t);
+            }
+            res.add(t);
+            cnt = num;
+        }
+        return res;
+    }
+}

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

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+# 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[List[int]]:
+        if root is None:
+            return []
+        q = Queue()
+        q.put(root)
+        cnt = 1
+        res = []
+        level = 0
+        while not q.empty():
+            level += 1
+            t = []
+            num = 0
+            for _ in range(cnt):
+                node = q.get()
+                t.append(node.val)
+                if node.left:
+                    q.put(node.left)
+                    num += 1
+                if node.right:
+                    q.put(node.right)
+                    num += 1
+            if (level & 1) == 0:
+                t.reverse()
+            res.append(t)
+            cnt = num
+        return res