题目

输入某二叉树的前序遍历和中序遍历的结果，请构建该二叉树并返回其根节点。

假设输入的前序遍历和中序遍历的结果中都不含重复的数字。

示例 1:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

示例 2:

Input: preorder = [-1], inorder = [-1]
Output: [-1]

限制：

0 <= 节点个数 <= 5000

解答

#include <iostream>
#include <unordered_map>
#include <vector>using namespace std;struct TreeNode
{int val;TreeNode *left;TreeNode *right;TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};class Solution {
public:TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {for (int i = 0; i < inorder.size(); i++){// 用中序遍历数组建立 值-下标 的映射value_index[inorder[i]] = i;}return recursive(preorder,0, 0, inorder.size() - 1);}TreeNode *recursive(vector<int>& pre,int pre_root, int in_left, int in_right){if (in_left > in_right) return nullptr;// 将对应的 val 赋给 node 节点TreeNode *node = new TreeNode(pre[pre_root]);int in_root = value_index[pre[pre_root]];node->left = recursive(pre,pre_root + 1, in_left, in_root - 1);node->right = recursive(pre,pre_root + in_root - in_left + 1, in_root + 1, in_right);return node;}
private:unordered_map<int, int> value_index;
};