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

public class Solution {
    TreeNode pre = null;
    
    public void flatten(TreeNode root) {
        if (root == null) return;
        helper(root);
    }
    
    public void helper(TreeNode cur) {
        if (pre != null) {
            pre.left = null;
            pre.right = cur;
        }
        pre = cur;
        TreeNode left = cur.left;
        TreeNode right = cur.right;
        if (left != null) helper(left);
        if (right != null) helper(right);
    }
}

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)。

public class Solution {
    public void flatten(TreeNode root) {
        ArrayList<TreeNode> pre = new ArrayList<TreeNode>();
        pre.add(root);
        helper(root, pre);
    }
    
    public void helper(TreeNode root, ArrayList<TreeNode> pre) {
        if (root == null) return;
        TreeNode left = root.left;
        TreeNode right = root.right;
        if (root != pre.get(0)) {
            TreeNode parent = pre.get(0);
            parent.left = null;
            parent.right = root;
            pre.set(0, root);
        }
        helper(left, pre);
        helper(right, pre);
    }
}

Iterative: 与平时写Preorder的iterative不大一样

public class Solution {
    public void flatten(TreeNode root) {
        if (root == null) return;
        TreeNode pre = null;
        Stack<TreeNode> st = new Stack<TreeNode>();
        TreeNode cur = root;
        st.push(cur);
        while (!st.isEmpty()) {
            cur = st.pop();
            if (pre != null) {
                pre.left = null;
                pre.right = cur;
            }
            pre = cur;
            if (cur.right != null) st.push(cur.right);
            if (cur.left != null) st.push(cur.left);
        }
    }
}