diff --git a/notes/面试总结.md b/notes/面试总结.md
index 6e4dfb66..1ca3aeb7 100644
--- a/notes/面试总结.md
+++ b/notes/面试总结.md
@@ -34,6 +34,8 @@
 * [29. 顺时针打印矩阵](#29-顺时针打印矩阵)
 * [30. 包含 min 函数的栈](#30-包含-min-函数的栈)
 * [31. 栈的压入、弹出序列](#31-栈的压入弹出序列)
+* [32.1 从上往下打印二叉树](#321-从上往下打印二叉树)
+* [32.2 把二叉树打印成多行](#322-把二叉树打印成多行)
 
 * [参考文献](#参考文献)
 <!-- GFM-TOC -->
@@ -2291,6 +2293,159 @@ class Solution:
             return False
         return True
 ```
+# 32.1 从上往下打印二叉树
+
+[NowCoder](https://www.nowcoder.com/practice/7fe2212963db4790b57431d9ed259701?tpId=13&tqId=11175&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
+
+## 题目描述
+
+从上往下打印出二叉树的每个节点，同层节点从左至右打印。
+
+例如，以下二叉树层次遍历的结果为：1,2,3,4,5,6,7
+
+<div align="center"> <img src="../pics//348bc2db-582e-4aca-9f88-38c40e9a0e69.png" width="250"/> </div><br>
+
+## 解题思路
+
+使用队列来进行层次遍历。
+
+不需要使用两个队列分别存储当前层的节点和下一层的节点，因为在开始遍历一层的节点时，当前队列中的节点数就是当前层的节点数，只要控制遍历这么多节点数，就能保证这次遍历的都是当前层的节点。
+
+```java
+public ArrayList<Integer> PrintFromTopToBottom(TreeNode root) {
+    Queue<TreeNode> queue = new LinkedList<>();
+    ArrayList<Integer> ret = new ArrayList<>();
+    queue.add(root);
+    while (!queue.isEmpty()) {
+        int cnt = queue.size();
+        while (cnt-- > 0) {
+            TreeNode t = queue.poll();
+            if (t == null)
+                continue;
+            ret.add(t.val);
+            queue.add(t.left);
+            queue.add(t.right);
+        }
+    }
+    return ret;
+}
+```
+
+```python
+# -*- coding:utf-8 -*-
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+class Solution:
+    # 返回从上到下每个节点值列表，例：[1,2,3]
+    def PrintFromTopToBottom(self, root):
+        # write code here
+        queue = []
+        ret = []
+        queue.append(root)
+        while queue:
+            cnt = len(queue)
+            while cnt > 0:
+                cnt -= 1
+                t = queue.pop(0)
+                if t == None:
+                    continue
+                ret.append(t.val)
+                queue.append(t.left)
+                queue.append(t.right)
+        return ret
+```
+
+```python
+# -*- coding:utf-8 -*-
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+class Solution:
+    # 返回从上到下每个节点值列表，例：[1,2,3]
+    def PrintFromTopToBottom(self, root):
+        # write code here
+        l=[]
+        if not root:
+            return []
+        q=[root]
+        while len(q):
+            t=q.pop(0)
+            l.append(t.val)
+            if t.left:
+                q.append(t.left)
+            if t.right:
+                q.append(t.right)
+        return l 
+```
+
+# 32.2 把二叉树打印成多行
+
+[NowCoder](https://www.nowcoder.com/practice/445c44d982d04483b04a54f298796288?tpId=13&tqId=11213&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
+
+## 题目描述
+
+和上题几乎一样。
+
+## 解题思路
+
+```java
+ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {
+    ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
+    Queue<TreeNode> queue = new LinkedList<>();
+    queue.add(pRoot);
+    while (!queue.isEmpty()) {
+        ArrayList<Integer> list = new ArrayList<>();
+        int cnt = queue.size();
+        while (cnt-- > 0) {
+            TreeNode node = queue.poll();
+            if (node == null)
+                continue;
+            list.add(node.val);
+            queue.add(node.left);
+            queue.add(node.right);
+        }
+        if (list.size() != 0)
+            ret.add(list);
+    }
+    return ret;
+}
+```
+
+```python
+# -*- coding:utf-8 -*-
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+class Solution:
+    # 返回二维列表[[1,2],[4,5]]
+    def Print(self, pRoot):
+        # write code here
+        ret = []
+        # queue = []
+        # queue.append(pRoot)
+        queue = [pRoot]
+        while queue:
+            list = []
+            cnt = len(queue)
+            while cnt > 0:
+                cnt -= 1
+                node = queue.pop(0)
+                if node == None:
+                    continue
+                list.append(node.val)
+                queue.append(node.left)
+                queue.append(node.right)
+            if len(list):
+                ret.append(list)
+        return ret
+```
 
 # 参考文献