题目

从前序与中序遍历序列构造二叉树
给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。

示例 1:

输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
输出: [3,9,20,null,null,15,7]
示例 2:

输入: preorder = [-1], inorder = [-1]
输出: [-1]

提示:

1 <= preorder.length <= 3000
inorder.length == preorder.length
-3000 <= preorder[i], inorder[i] <= 3000
preorder 和 inorder 均无重复元素
inorder 均出现在 preorder
preorder 保证为二叉树的前序遍历序列
inorder 保证为二叉树的中序遍历序列
相关标签

C++

答案

/**

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:
unordered_map<int,int> inorder_map;
TreeNode *buildTreeHelper(vector&preorder,vector &inorder,int start,int end,int &pre_index){
if(start > end){
return nullptr;
}
int inorder_index = inorder_map[preorder[pre_index]];
TreeNode *rootnode = new TreeNode(preorder[pre_index++]);

rootnode->left = buildTreeHelper(preorder,inorder,start,inorder_index-1,pre_index);
rootnode->right = buildTreeHelper(preorder,inorder,inorder_index+1,end,pre_index);
return rootnode;
}

TreeNode *buildTree(vector &preorder, vector &inorder)
{

for(int i=0;i<inorder.size();i++){
    inorder_map[inorder[i]] = i;
}
int pre_index = 0;
return buildTreeHelper(preorder,inorder,0,inorder.size()-1,pre_index);

}

};