一.题目

Construct Binary Tree from Preorder and Inorder Traversal

Total Accepted: 36475 Total Submissions: 138308My
Submissions

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:

You may assume that duplicates do not exist in the tree.

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二.解题技巧

这道题仅仅是考察先序和中序遍历的概念，先序是先訪问根节点，然后訪问左子树。最后訪问右子树；中序遍历是先遍历左子树，然后訪问根节点。最后訪问右子树。

做法都是先依据先序遍历的概念，找到先序遍历的第一个值，即为根节点的值。然后依据根节点将中序遍历的结果分成左子树和右子树。然后就能够递归的实现了。

上述做法的时间复杂度为O(n^2)。空间复杂度为O(1)

三.实现代码

#include <iostream>
#include <algorithm>
#include <vector>

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

using std::vector;
using std::find;

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.体会

这道题是考察基础概念的题。并不须要非常多算法，仅仅是一个递归的过程。

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