From 45af5df13e720dd1ffa1c47ce6a7ee9460e256e2 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <14021051@buaa.edu.cn>
Date: Fri, 2 Nov 2018 23:13:14 +0800
Subject: [PATCH] Create 107. Binary Tree Level Order Traversal II.md

---
 107. Binary Tree Level Order Traversal II.md | 46 ++++++++++++++++++++
 1 file changed, 46 insertions(+)
 create mode 100644 107. Binary Tree Level Order Traversal II.md

diff --git a/107. Binary Tree Level Order Traversal II.md b/107. Binary Tree Level Order Traversal II.md
new file mode 100644
index 0000000..87cd8de
--- /dev/null
+++ b/107. Binary Tree Level Order Traversal II.md	
@@ -0,0 +1,46 @@
+# [107. Binary Tree Level Order Traversal II](https://leetcode.com/problems/binary-tree-level-order-traversal-ii/description/)
+# 思路
+题目的意思就是层序遍历的变形，要逆序输出每一层的节点。为此我们先正常层序遍历并用一个stack记录每一层的节点。然后再依次出栈即可。     
+用last指针指向每一层的最后一个节点，每当遍历到这个节点即说明遍历完一层，此时应该将此层所有节点压入栈。last初始为root，
+后面每当遍历完每层最后一个节点后，即将last更新成下一层的最后一个节点，为此需要用一个next_last来不断记录能确定的下一层的最右节点。   
+时间复杂度和空间复杂度都是O(n)
+
+# C++
+``` C++
+class Solution {
+public:
+    vector<vector<int>> levelOrderBottom(TreeNode* root) {
+        TreeNode *p, *next_last,*last=root;
+        stack<vector<int>>stk;
+        queue<TreeNode *>q;
+        vector<vector<int>>res;
+        if(root == NULL) return res;
+        vector<int>tmp; // 存放一层的节点
+        q.push(root);
+        while(!q.empty()){
+            p = q.front();
+            q.pop();
+            tmp.push_back(p -> val);
+            if(p -> left){
+                q.push(p -> left);
+                next_last = p -> left;
+            }
+            if(p -> right) {
+                q.push(p -> right);
+                next_last = p -> right;
+            }
+            if(p == last){
+                stk.push(tmp);
+                tmp.clear();
+                last = next_last; // 更新last指针
+            }
+        }
+        while(!stk.empty()){
+            res.push_back(stk.top());
+            stk.pop();
+        }
+        return res;
+    }
+};
+```
+