LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal

1. 题目描述

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

2. 解题思路

这是一道数据结构的题目，给出二叉树的前序和中序遍历结果，从而还原二叉树。

这个题目我们用递归的方式去思考：

1. 前序遍历的第一个点，肯定是根节点。

2. 利用前序遍历得到的根节点在中序遍历的数组中找到那个点，从而，在那个点前面的就是左子树，后面的就是右子树。

3.对左右子树继续以上过程，那么就可以构造出二叉树了。

3. 实现代码

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:     TreeNode* buildTree(vector<int> &preorder, vector<int> &inorder) {        return Helper(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);    }        TreeNode* Helper(vector<int> &preorder, int begin1, int end1, vector<int> &inorder, int begin2, int end2)    {        if(begin1 > end1)            return NULL;        else if(begin1 == end1)            return new TreeNode(preorder[begin1]);                    TreeNode* root = new TreeNode(preorder[begin1]);        int i = begin2;        for(; i <= end2; i ++)        {            if(inorder[i] == preorder[begin1])                break;        }        //inorder[i] is the root        int leftlen = i-begin2;                //preorder[begin1] is root        //inorder[begin2+leftlen] is root        root->left = Helper(preorder, begin1+1, begin1+leftlen, inorder, begin2, begin2+leftlen-1);        root->right = Helper(preorder, begin1+leftlen+1, end1, inorder, begin2+leftlen+1, end2);        return root;    }};