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。