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.empty() || vinOrder.empty()){
            return nullptr;
        }
        auto node = new TreeNode(preOrder[0]);
        if(preOrder.size() == 1){
            return node;
        }
        auto it = find(vinOrder.begin(), vinOrder.end(),preOrder[0]);
        if(it != vinOrder.end()){
            vector<int> left_inorder(vinOrder.begin(), it);
            vector<int> right_inorder(it + 1, vinOrder.end());
            int nleft = left_inorder.size();
            int nright = right_inorder.size();
            vector<int> left_preorder(preOrder.begin()+1,preOrder.begin()+nleft+1);
            vector<int> right_preorder(preOrder.begin() + nleft + 1, preOrder.end());
            node ->left = reConstructBinaryTree(left_preorder, left_inorder);
            node ->right = reConstructBinaryTree(right_preorder, right_inorder);
        }
        return node;
    }
};