leetcode笔记：Construct Binary Tree from Preorder and Inorder Traversal

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。