[leetcode] Flatten Binary Tree to Linked List

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
思路：
先序遍历的变形，有两种方法。一种从上往下，一种从下往上。
先说从上往下。从根节点开始，将当前节点的坐节点移到右节点上，原来的右节点与现在的最右节点相连。这样，当前节点的左边就为NULL，并且它的右节点就是先序遍历接下来应该访问的值。
在说从下往上。对某个节点进行操作，先假设右边全部有序（所以先递归右节点），其余的和从上往下一样，只是这种方法先递归到树中最右节点，再开始操作。
题解：

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

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