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 <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
        int n = preOrder.size(), m = vinOrder.size();
        if(n == 0 || m == 0) return nullptr;
        TreeNode* root = new TreeNode(preOrder[0]);
        for(int i = 0; i < m; i ++)
        {
            if(vinOrder[i] == preOrder[0])
            {
                vector<int> leftpre(preOrder.begin() + 1, preOrder.begin() + i + 1);
                vector<int> leftvin(vinOrder.begin(), vinOrder.begin() + i);
                root->left = reConstructBinaryTree(leftpre, leftvin);

                vector<int> rightpre(preOrder.begin() + i + 1, preOrder.end());
                vector<int> rightvin(vinOrder.begin() + i + 1, vinOrder.end());
                root->right = reConstructBinaryTree(rightpre, rightvin);
                break;
            }
        }
        return root;
    }
};