Flatten Binary Tree to Linked List

Problem Description:

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.

思路一，按照题意从根节点开始，如果当前节点有左孩子，就将它的左孩子添加到自己和右孩子之间，这里每次需要找到左孩子最右边的节点，连接到当前右孩子。然后依次往右处理自己的右孩子，直到右孩子为空。具体代码如下：

/**
 * Definition for binary tree
 * 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==NULL)
            return;
        while (root != NULL) 
        {
			if (root->left != NULL) 
			{
				TreeNode *p = root->left;
				while (p->right != NULL) //找到左孩子的最右边节点
				{
					p = p->right;
				}
				p->right = root->right;
				root->right = root->left;
				root->left = NULL;
			}
			root = root->right;
		}
    }
};

思路二：可以看出变换后的树实际上是按照先序遍历的顺序排列的，因此可以利用中序递归遍历，记录先序遍历的前驱结点，依次调整每个孩子的左右子树，这里需要注意的是遍历过程中需要先将当前节点的右子树记录下来，再调整当前节点的左右子树，然后递归调整左右子树。具体代码如下：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:

    TreeNode *pre=NULL;

    void flatten(TreeNode *root) {
        if(root==NULL)
            return;
        TreeNode *lastright=root->right;//记录当前节点的右子树
        if(pre)
        {
            pre->left=NULL;
            pre->right=root;
        }
        pre=root;
        flatten(root->left);
        flatten(lastright);
    }
};