LeetCode刷题_c++版-114二叉树的展开为链表

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
    
    
public:
    void preOrder(TreeNode*& node, TreeNode*& last){
    
    
        if(node == NULL) return;
        if(node->left == NULL && node->right == NULL){
    
    
            last = node;
            return;
        }
        //节点备份
        TreeNode* left = node->left;
        TreeNode* right = node->right;
        //左子树放到根节点的右子树上，左指针置为空
        if(left){
    
    
            node->right = left;
            node->left = NULL;
            last = left;
            preOrder(left, last);
        }
        //右子树放到最后一个操作的节点的右子树上
        if(right){
    
    
            last->right = right;
            last = right;
            preOrder(right, last);
        }
        return;
    }    
    
    void flatten(TreeNode* root) {
    
    
        if(root == NULL || (root->left == NULL && root->right == NULL)) return;
        //last记录最后一个操作的节点
        TreeNode* last = root;
        preOrder(root, last);
        return;
    }


};