一.题目
Construct Binary Tree from Preorder and Inorder Traversal
Total Accepted: 36475 Total Submissions: 138308 My SubmissionsGiven 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
这道题只是考察先序和中序遍历的概念,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。
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,并不需要很多算法,只是一个递归的过程。
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