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
        if (preOrder.size() == 0 || vinOrder.size() == 0) {
            return nullptr;
        }

        TreeNode* root = new TreeNode(preOrder[0]);

        for (int i = 0; i < vinOrder.size(); i++) {
            if (vinOrder[i] == preOrder[0]) {
                vector<int> leftVinOrderTree(vinOrder.begin(), vinOrder.begin() + i);
                vector<int> leftPreOrderTree(preOrder.begin() + 1,
                                             preOrder.begin() + leftVinOrderTree.size() + 1);
                root->left = reConstructBinaryTree(leftPreOrderTree, leftVinOrderTree);
                vector<int> rightVinOrderTree(vinOrder.begin()+i+1, vinOrder.end());
                vector<int> rightPreOrderTree(preOrder.begin()+i+1,
                                             preOrder.end());
                root->right = reConstructBinaryTree(rightPreOrderTree, rightVinOrderTree);
            }
        }

        return root;
    }

    vector<int> copyTree(vector<int> father, vector<int>::iterator begin,
                         vector<int>::iterator end) {
        vector<int> son;
        vector<int>::iterator it;
        for (it = begin; it != end; it++) {

        }

        return son;
    }
};