[Leetcode]Flatten Binary Tree to Linked List~

Flatten Binary Tree to Linked List My Submissions Question
Total Accepted: 66064 Total Submissions: 223714 Difficulty: Medium
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

click to show hints.

Hints:
If you notice carefully in the flattened tree, each node’s right child points to the next node of a pre-order traversal.

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中午上课前在leetcode做的时候出了各种细节错误，于是上课的时候在纸上写了下，也就十多分钟。上完课我回来提交，AC。
有时候在纸上写会更细心，写的过程也在思考。
思路：
递归处理，将左右子树Flatten，并判断左子树是否为空，为空则直接返回。不为空则将左子树的最后一个结点的右孩子设为右子树，并将左子树设置为根结点的右子树，并清空根节点左子树即可

/**
 * 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) {
        if(!root)   return;
        if(!root->left && !root->right) return;
        flatten(root->left);
        flatten(root->right);
        if(end(root->left)){
            end(root->left) -> right = root->right;
            root->right = root->left;
            root->left = nullptr;
            return;
        }
        else{
            return;
        }
    }
    TreeNode* end(TreeNode* root){
        if(!root)   return nullptr;
        auto p = root;
        while(p->right){
            p = p->right;
        }
        return p;
    }
};

P.S 上张图记录下，这种感觉蛮好的