[LeetCode] Flatten Binary Tree to Linked List

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

For example, given the following tree:

    1
   / 
  2   5
 /    
3   4   6

The flattened tree should look like:

1
 
  2
   
    3
     
      4
       
        5
         
          6

将一个二叉树重组成一个链表

题目要求按照二叉树的先序遍历的顺序重组二叉树

思路1：

找到最左侧结点，将其父结点与父结点右结点断开，将其连接至父结点右侧，变成父结点的右结点，然后把原右结点插入到新右结点的右侧。递归这一操作即可

/**
 * 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)
            flatten(root->left);
        if (root->right)
            flatten(root->right);
        TreeNode* temp = root->right;
        root->right = root->left;
        root->left = NULL;
        while (root->right)
            root = root->right;
        root->right = temp;
    }
};

思路2：

从根结点出发，判断其左结点是否存在，如果存在，则将根结点与其右结点断开，根的左结点变成其右结点。在将右结点链接至左结点最右边的右结点处。

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