From 5bc0d437f0086b020165cfee527813cddc5e2db3 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: Fri, 26 Apr 2019 23:00:45 +0800
Subject: [PATCH] Create 105. Construct Binary Tree from Preorder and Inorder
 Traversal.md

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 create mode 100644 solutions/105. Construct Binary Tree from Preorder and Inorder Traversal.md

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+# [105. Construct Binary Tree from Preorder and Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)
+# 思路
+根据中序和前序遍历建树。        
+由于先序遍历的第一个肯定是根，所以原二叉树的根节点可以知道，然后由于树中没有相同元素，所以我们可以在中序遍历中也定位出根节点的位置，
+并以根节点的位置将中序遍历拆分为左右两个部分，分别对其递归调用原函数。其中在中序遍历中定位根节点可以用个for循环也可以通过map。
+
+需要提一点的是，同时知道前序遍历（或后序遍历）和中序遍历就能够唯一确定一颗二叉树，而前序和后序则不能。
+
+# C++
+``` C++
+class Solution {
+private:
+    TreeNode* helper(int num, vector<int>& preorder, int pre_start, 
+                     vector<int>& inorder, int in_start, 
+                     map<int, int>& mp){
+        if(num == 0) return NULL;
+        int val = preorder[pre_start];
+        TreeNode *node = new TreeNode(val);
+        int i = mp[val] - in_start;
+        node -> left = helper(i, preorder, pre_start + 1, inorder, in_start, mp);
+        node -> right = helper(num-i-1, preorder, pre_start+i+1, inorder, 
+                               in_start+i+1, mp);
+        return node;
+    }
+public:
+    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
+        map<int, int>mp;
+        // 用map记录某个节点在中序遍历数组中的索引
+        for(int i = 0; i < inorder.size(); i++) mp[inorder[i]] = i;
+        return helper(preorder.size(), preorder, 0, inorder, 0, mp);   
+    }
+};
+```