leetcode Question 30: 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

Analysis:

Note that the problem requires in-place operation.The flatten procedure is like:  cut the left child and set to right, the right child is then linked to somewhere behind the left child. Where should it be then?  Actually the right child should be linked to the most-right node of the left node. So the algorithm is as follows:

(1) store the right child (we call R)

(2) find the right-most node of left child

(3) set R as the right-most node's right child.

(4) set left child as the right child

(5) set the left child NULL

(6) set current node to current node's right child.

(7) iterate these steps until all node are flattened.

Code(Python):

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

Code(Python):

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param root, a tree node
    # @return nothing, do it in place
    def flatten(self, root):
        while root != None:
            if root.left == None:
                root = root.right
            else:
                tmp = root.left
                while tmp.right != None:
                    tmp = tmp.right
                tmp.right = root.right
                root.right = root.left
                root.left = None