Leetcode: 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
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.

第二遍做法：不用ArrayList来wrap up previous node, 直接把previous node做成全局变量

 1 public class Solution {
 2     TreeNode pre = null;
 3     
 4     public void flatten(TreeNode root) {
 5         if (root == null) return;
 6         helper(root);
 7     }
 8     
 9     public void helper(TreeNode cur) {
10         if (pre != null) {
11             pre.left = null;
12             pre.right = cur;
13         }
14         pre = cur;
15         TreeNode left = cur.left;
16         TreeNode right = cur.right;
17         if (left != null) helper(left);
18         if (right != null) helper(right);
19     }
20 }

第一遍做法：用递归来解决，维护先序遍历的前一个结点pre，然后每次把pre的左结点置空，右结点设为当前结点。这里需要注意的一个问题就是我们要先把左右子结点（Code Ganker说只需要右节点，我是保存了左右节点）保存一下，以便等会可以进行递归，否则有可能当前结点的右结点会被覆盖，后面就取不到了。算法的复杂度时间上还是一次遍历，O(n)。空间上是栈的大小，O(logn)。

 1 public class Solution {
 2     public void flatten(TreeNode root) {
 3         ArrayList<TreeNode> pre = new ArrayList<TreeNode>();
 4         pre.add(root);
 5         helper(root, pre);
 6     }
 7     
 8     public void helper(TreeNode root, ArrayList<TreeNode> pre) {
 9         if (root == null) return;
10         TreeNode left = root.left;
11         TreeNode right = root.right;
12         if (root != pre.get(0)) {
13             TreeNode parent = pre.get(0);
14             parent.left = null;
15             parent.right = root;
16             pre.set(0, root);
17         }
18         helper(left, pre);
19         helper(right, pre);
20     }
21 }

Iterative: 与平时写Preorder的iterative不大一样

 1 public class Solution {
 2     public void flatten(TreeNode root) {
 3         if (root == null) return;
 4         TreeNode pre = null;
 5         Stack<TreeNode> st = new Stack<TreeNode>();
 6         TreeNode cur = root;
 7         st.push(cur);
 8         while (!st.isEmpty()) {
 9             cur = st.pop();
10             if (pre != null) {
11                 pre.left = null;
12                 pre.right = cur;
13             }
14             pre = cur;
15             if (cur.right != null) st.push(cur.right);
16             if (cur.left != null) st.push(cur.left);
17         }
18     }
19 }