## The inverse of a tree with root r, and subtrees right and left, is a tree with root r, whose right subtree is the inverse of left, and whose left subtree is the inverse of right.

### For example:

  For example  Given binary tree         1     /   \    2     3   / \   / \  4   5 6   7  
invert and return         1     /   \    3     2   / \   / \  7   6 5   4  

public TreeNode invertTree(TreeNode root) {      if (root == null) {          return null;      }      TreeNode right = invertTree(root.right);      TreeNode left = invertTree(root.left);      root.left = right;      root.right = left;      return root;  }

### Complexity Analysis

Since each node in the tree is visited only once, the time complexity is O(n), where nn 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.

Because of recursion, O(h)O(h) function calls will be placed on the stack in the worst case, where hh is the height of the tree. Because h\in O(n)h∈O(n), the space complexity is O(n).

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Posted By Blogger to HDGEM at 3/04/2017 07:36:00 AM