**一. 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. **二. 题目分析** 这道题考察了先序和中序遍历，先序是先访问根节点，然后访问左子树，最后访问右子树；中序遍历是先遍历左子树，然后访问根节点，最后访问右子树。 做法都是先根据先序遍历的概念，找到先序遍历的第一个值，即为根节点的值，然后根据根节点将中序遍历的结果分成左子树和右子树，然后就可以递归的实现了。 按照上述做法，时间复杂度为`O(n^2)`，空间复杂度为`O(1)` **三. 示例代码** ~~~ #include <iostream> #include <algorithm> #include <vector> using namespace std; struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; class Solution { private: TreeNode* buildTree(vector<int>::iterator PreBegin, vector<int>::iterator PreEnd, vector<int>::iterator InBegin, vector<int>::iterator InEnd) { if (PreBegin == PreEnd) { return NULL; } int HeadValue = *PreBegin; TreeNode *HeadNode = new TreeNode(HeadValue); vector<int>::iterator LeftEnd = find(InBegin, InEnd, HeadValue); if (LeftEnd != InEnd) { HeadNode->left = buildTree(PreBegin + 1, PreBegin + (LeftEnd - InBegin) + 1, InBegin, LeftEnd); } HeadNode->right = buildTree(PreBegin + (LeftEnd - InBegin) + 1, PreEnd, LeftEnd + 1, InEnd); return HeadNode; } public: TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) { if (preorder.empty()) { return NULL; } return buildTree(preorder.begin(), preorder.end(), inorder.begin(), inorder.end()); } }; ~~~ **四. 小结** 该题考察了基础概念，并不涉及过多的算法问题。