From c1ce0a4ed47bf2a6fcc60b3268c3d2438a77c000 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <tangshusen@pku.edu.cn>
Date: Wed, 3 Apr 2019 23:22:27 +0800
Subject: [PATCH] Create 94. Binary Tree Inorder Traversal.md

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+# [94. Binary Tree Inorder Traversal](https://leetcode.com/problems/binary-tree-inorder-traversal/)
+# 思路
+中序遍历二叉树。属于数据结构基本题。
+## 思路一、递归
+二叉树的中序、前序、后序遍历最方便的当然就是使用递归了。中序遍历某个节点时，先不访问它，先递归遍历其左子树，然后才访问该节点，最后递归遍历其右节点。
+
+## 思路二、非递归
+我们先定义一个栈，然后从根节点开始，进入循环，若当前节点不空那么将其入栈（暂不访问），然后向左下降进入下一循环；若为空则说明往左下降不动了，应该将栈顶元素取出
+并访问，然后向右下降；循环直到栈空且当前节点也为空为止。
+
+# C++
+## 思路一
+``` C++
+class Solution {
+private:
+    void helper(TreeNode* root, vector<int> &res){
+        if(!root) return;
+        helper(root -> left, res);
+        res.push_back(root -> val);
+        helper(root -> right, res);
+        
+    }
+public:
+    vector<int> inorderTraversal(TreeNode* root) {
+        vector<int>res;
+        helper(root, res);
+        return res;
+    }
+};
+```
+
+## 思路二
+``` C++
+class Solution {
+public:
+    vector<int> inorderTraversal(TreeNode* root) {
+        vector<int>res;
+        if(!root) return res;
+        stack<TreeNode *>stk;
+        TreeNode* p = root;        
+        while(!stk.empty() || p){
+            if(p){
+                stk.push(p);
+                p = p -> left; // 一路向左
+            }
+            else{
+                p = stk.top();
+                stk.pop();
+                res.push_back(p -> val);
+                p = p -> right;
+            }
+        }
+        return res;
+    }
+};
+```