题解 | #重建二叉树#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
#include <algorithm>
#include <iterator>
#include <vector>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param preOrder int整型vector 
     * @param vinOrder int整型vector 
     * @return TreeNode类
     */
    TreeNode* reConstructBinaryTree(vector<int>& preOrder, vector<int>& vinOrder) {
        // write code here
        if (preOrder.size()==0||vinOrder.size()==0) {
            return nullptr;
        }
        TreeNode * tree=new TreeNode(preOrder[0]);
        int mid=distance(begin(vinOrder), find(vinOrder.begin(), vinOrder.end(), preOrder[0]));
        vector<int> preleft(preOrder.begin() + 1,preOrder.begin() + 1+ mid);
        vector<int> preright(preOrder.begin()+mid+1,preOrder.end());
        vector<int> left_in(vinOrder.begin(), vinOrder.begin() + mid);
        vector<int> right_in(vinOrder.begin() + mid + 1, vinOrder.end());
        tree->left=reConstructBinaryTree(preleft, left_in);
        tree->right=reConstructBinaryTree(preright, right_in);
        return tree;
    }
};