Binary Tree Upside Down

Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.

For example:
Given a binary tree {1,2,3,4,5},

    1
   / 
  2   3
 / 
4   5

return the root of the binary tree [4,5,2,#,#,3,1].

   4
  / 
 5   2
    / 
   3   1  

将left node 入stack， 然后将每个原来的left node的 left = parent.right, right = parent. (从下往上做)

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode UpsideDownBinaryTree(TreeNode root) {
        if(root == null) return root;
        LinkedList<TreeNode> ll = new LinkedList<TreeNode> ();
        ll.push(root);
        while(root.left != null){
            ll.push(root.left);
            root = root.left;
        }
        while(!ll.isEmpty()){
            TreeNode tmp = ll.pop();
            tmp.right = ll.peek();
            tmp.left = ll.peek() == null ? null : ll.peek().right;
        }
        return root;
    }
}

 从上往下做：

public class Solution {
    public TreeNode UpsideDownBinaryTree(TreeNode root) {
        if(root == null) return root;
        TreeNode cur = root, last = null, lastRight = null;
        while(cur != null){
            TreeNode tmp = cur.left;
            cur.left = lastRight;
            lastRight = cur.right;
            cur.right = last;
            last = cur;
            cur = tmp;
        }
        return last;
    }
}