[LeetCode 156] 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

solution:

use stack to store left node from top to bottom, then reverse to construct the tree.

if(root ==null || (root.left == null && root.right == null)) return root;
        Stack<TreeNode> path = new Stack<>();
        while(root.left!=null) {
            path.add(root);
            root = root.left;
        }
        path.add(root);
        while(!path.isEmpty()) {
            TreeNode temp = path.pop();
            if(!path.isEmpty()) {
                temp.right = path.peek();
                temp.left = path.peek().right;
            }else{
                temp.right = null;
                temp.left = null;
            }
        }
        return root;