Invert Binary Tree

Invert a binary tree.

Example:

Input:

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

Output:

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

Time complexity : O(n). Since each node in the tree is visited only once, the time complexity is O(n), where n is the number of nodes in the tree. We cannot do better than that, since at the very least we have to visit each node to invert it.
Space complexity : O(h) where h is height, or O(log(n)) - best, o(n) worst. O(h) function calls will be placed on the stack in the worst case, where h is the height of the tree. Because h ∈ O(n), the space complexity is O(n) (worst case). The best case is for a completely balanced tree. The worst case is for an unbalanced tree.

Look at solution 2 in leetcode for a queue based approach with the same time and space complexities