[LeetCode]Construct Binary Tree from Preorder and Inorder Traversal

Construct Binary Tree from Preorder and Inorder Traversal

 

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.

分析:

1、先序遍历第一个元素是该树的根

2、中序遍历该元素将树分为左右两个子树，该元素设为i

3、中序0~i-1对应于先序的1~i元素，中序右子树i+1~in_end对应于先序i+1~pre_end

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */

//递归,时间复杂度 O(n),空间复杂度O(logn)

class Solution {public:    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {        return build(begin(preorder),end(preorder),begin(inorder),end(inorder));    }        template<typename InputIterator>        TreeNode* build(InputIterator pre_first, InputIterator pre_last, InputIterator in_first, InputIterator in_last)    {        if(pre_first == pre_last)        {            return nullptr;        }        if(in_first == in_last)        {            return nullptr;        }                auto root = new TreeNode(*pre_first);        auto inRootPos = find(in_first, in_last, *pre_first);        auto leftSize = distance(in_first, inRootPos);                root->left = build(next(pre_first),next(pre_first,leftSize+1),in_first,next(in_first,leftSize));        root->right = build(next(pre_first,leftSize+1),pre_last,next(inRootPos),in_last);        return root;    }    };