Construct Binary Tree from Preorder and Inorder Traversal
最后更新于：2022-04-01 22:57:42
**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
#include
#include
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::iterator PreBegin, vector::iterator PreEnd,
vector::iterator InBegin, vector::iterator InEnd)
{
if (PreBegin == PreEnd)
{
return NULL;
}
int HeadValue = *PreBegin;
TreeNode *HeadNode = new TreeNode(HeadValue);
vector::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& preorder, vector& 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。
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