Leetcode 114. Flatten Binary Tree to Linked List

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

         1
        / 
       2   5
      /    
     3   4   6

The flattened tree should look like:

   1
    
     2
      
       3
        
         4
          
           5
            
             6

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void flatten(TreeNode* root) {
       TreeNode *pre = nullptr; //记住前一结点
       helper(root,pre);        //前序遍历
    }
    void helper(TreeNode *root, TreeNode *&pre) // 注意这里是一个指针的引用，pre是全局变量，不断变化
    {
        if (root == nullptr) return;
        if (pre != nullptr)
        {
            pre -> left = nullptr; // 将前一结点的左指针置空
            pre -> right = root;   // 将前一结点的右指针指向当前的根节点
        }
        pre = root;
        TreeNode *left = root -> left;
        TreeNode *right = root -> right;
        if (left != nullptr)   // 对左子树进行遍历
            helper(left,pre);  
        if (right != nullptr)  // 对右子树进行遍历
            helper(right,pre);
    }
};