105. Construct Binary Tree from Preorder and Inorder Traversal

1. 问题描述
   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. 解决思路
   利用递归建树的思想，找到左右子树的list，然后递归下去，就可以了。

3. 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,inorder,0,preorder.size()-1,0,inorder.size()-1);    }    TreeNode* helper(vector<int>& preorder, vector<int>& inorder, int pre_begin, int pre_end, int in_begin, int in_end) {        if (pre_end < pre_begin)            return NULL;        TreeNode *root = new TreeNode(preorder[pre_begin]);        int left_len = 0;        for (int i = in_begin; i <=in_end; ++i) {            if (inorder[i] == preorder[pre_begin])                break;            ++left_len;        }        root->left = helper(preorder,inorder,pre_begin+1,pre_begin+left_len,in_begin,in_begin+left_len-1);        root->right = helper(preorder,inorder,pre_begin+left_len+1,pre_end,in_begin+left_len+1,in_end);        return root;    }};