题目:

思想:
首先先序遍历能确定根节点的值,此时查看该值在中序遍历中的位置(如果索引为i),那么i左侧为左子树,i 右侧为右子树。从中序数组中即可看出左子树结点个数为 i,右子树节点个数为inorder.size()-i-1。也就代表先序数组中除了第一个元素外,先 i 个元素是左子树对应的先序数组元素,后面的元素为右子树对应的先序数组元素。递归的形式就出现啦!如果没想到可以看一下函数的参数。
代码如下:
C++:
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if(preorder.size()==0){return nullptr;}
        TreeNode* head=new TreeNode;
        head->val=preorder[0];
        int idx=find(inorder.begin(),inorder.end(),preorder[0])-inorder.begin();
        //preorder---左子树
        vector<int>::const_iterator Firstprel=preorder.begin()+1;
        vector<int>::const_iterator Secondprel=preorder.begin()+idx+1;
        vector<int> prel;
        prel.assign(Firstprel,Secondprel);
        //preorder---右子树
        vector<int>::const_iterator Firstprer=preorder.begin()+idx+1;
        vector<int>::const_iterator Secondprer=preorder.end();
        vector<int> prer;
        prer.assign(Firstprer,Secondprer);
        //inorder---左子树
        vector<int>::const_iterator Firstinl=inorder.begin();
        vector<int>::const_iterator Secondinl=inorder.begin()+idx;
        vector<int> inl;
        inl.assign(Firstinl,Secondinl);
        //inorder---右子树
        vector<int>::const_iterator Firstinr=inorder.begin()+idx+1;
        vector<int>::const_iterator Secondinr=inorder.end();
        vector<int> inr;
        inr.assign(Firstinr,Secondinr);
        head->left=buildTree(prel,inl);
        head->right=buildTree(prer,inr);
        return head;
    }
};注意看一下这里的写法:
int idx=find(inorder.begin(),inorder.end(),preorder[0])-inorder.begin();参考博文:(在inorder中寻找preorder[0]这个元素,返回其索引值)
c++vector查找元素所在的索引下标_vector查找元素索引-CSDN博客
//preorder---左子树
vector<int>::const_iterator Firstprel=preorder.begin()+1;
vector<int>::const_iterator Secondprel=preorder.begin()+idx+1;
vector<int> prel;
prel.assign(Firstprel,Secondprel);参考博文:(就是把preorder中的第一个元素直到第idx个元素,复制给prel)
vector 切片,截取指定区间元素_vector截取-CSDN博客
Python:
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
        if len(preorder)==0:
            return None
        head=TreeNode(preorder[0])
        idx=inorder.index(preorder[0])
        prel=preorder[1:idx+1]
        prer=preorder[idx+1:]
        inl=inorder[:idx+1]
        inr=inorder[idx+1:]
        head.left=self.buildTree(prel,inl)
        head.right=self.buildTree(prer,inr)
        return head
        注意这样几个写法:
Python中的空指针为None
Python中的创建实例
head=TreeNode(preorder[0])Python中的从inorder中寻找preorder[0],并返回索引
idx=inorder.index(preorder[0])


















