Invert Binary Tree

Invert a binary tree.

     4
   /   \
  2     7
 / \   / \
1   3 6   9

to

     4
   /   \
  7     2
 / \   / \
9   6 3   1

Solution1: recursive
Time complexity is O(n), space complexity is stack space (O(logn)).

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return root;
        }
        
        TreeNode temp = root.left;
        root.left = root.right;
        root.right= temp;
        
        invertTree(root.left);
        invertTree(root.right);
        
        return root;
    }
}

Solution2: iterative
Time complexity is O(n), space complexity is O(n).

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return root;
        }
        
        LinkedList queue = new LinkedList();
        queue.add(root);
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            if (node.left != null) {
                queue.add(node.left);
            }
            if (node.right != null) {
                queue.add(node.right);
            }
            
            TreeNode temp = node.left;
            node.left = node.right;
            node.right = temp;
        }
        return root;
    }
}