156 Binary Tree Upside Down

156. Binary Tree Upside Down

1. Question

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

2. Implementation

(1) Recursion

class Solution {
    public TreeNode upsideDownBinaryTree(TreeNode root) {
        if (root == null || root.left == null) {
            return root;
        }

        TreeNode newRoot = upsideDownBinaryTree(root.left);
        root.left.left = root.right;
        root.left.right = root;
        root.left = null;
        root.right = null;
        return newRoot;
    }
}

(2) Iteration

class Solution {
    public TreeNode upsideDownBinaryTree(TreeNode root) {
        TreeNode curNode = root, preNode = null, nextNode = null, tempNode = null;

        while (curNode != null) {
            nextNode = curNode.left;

            curNode.left = tempNode;
            tempNode = curNode.right;
            curNode.right = preNode;
            preNode = curNode;
            curNode = nextNode;
        }
        return preNode;
    }
}

3. Time & Space Complexity

Recursion: 时间复杂度O(h), 空间复杂度O(h)

Iteration: 时间复杂度O(h), 空间复杂度O(1)