LeetCode 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

感觉以前写过blog了，先来个递归版本（注意left指针全部要设置成NULL）

/**
 * 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 {
private:
    TreeNode* last;
public:
    void flatten(TreeNode* root) {
        last = NULL;
        dfs(root);
    }
    
    void dfs(TreeNode* root) {
        if (root == NULL) {
            return;
        }
        
        TreeNode* r = root->right;
        TreeNode* l = root->left;
        
        if (last != NULL) {
            last->right = root;
        }

        last = root;
        last->left = NULL;

        dfs(l);
        dfs(r);
    }
};

非递归版本discuss里找的。

/**
 * 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) {
        while (root) {
            if (root->left && root->right) {
                TreeNode* largest = root->left;
                while (largest->right) {
                    largest = largest->right;
                }
                largest->right = root->right;
            }
            if (root->left) {
                root->right = root->left;
            }
            root->left = NULL;
            root = root->right;
        }
    }
};