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

Thoughts:

1. Post order (right first); + record prev node

2. d

Code: Recursive

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private TreeNode prev = null;
    public void flatten(TreeNode root) {
        if (root == null) return;
        flatten(root.right);
        flatten(root.left);
        root.right = prev;
        root.left = null;
        prev = root;
    }
}

Code: Iterative: Using stack

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

class Solution(object):
    def flatten(self, root):
        """
        :type root: TreeNode
        :rtype: void Do not return anything, modify root in-place instead.
        """
        if not root: return
        stack = [root]
        while stack:
            cur = stack.pop()
            if cur.right:
                stack.append(cur.right)
            if cur.left:
                stack.append(cur.left)
            if stack:
                cur.right = stack[-1]
            cur.left = None

Code: Iterative Traversal C++:

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

            top = top -> right;
        }
    }
};