# 226 Invert Binary Tree – Easy

### Problem:

Invert a binary tree.

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

to

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

### Thoughts:

This problem is very straightforward, I come up with two version, one is using recursion and the other is using iterative which is BFS.

### Solutions:

``````/**
* 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) {
return recursion(root);
// return iterative(root);
}
private TreeNode recursion(TreeNode root) {
if (root == null) {
return null;
}
TreeNode tmp = root.left;
root.left = root.right;
root.right = tmp;
recursion(root.left);
recursion(root.right);
return root;
}
private TreeNode iterative(TreeNode root) {
if (root == null) {
return null;
}
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.add(root);
while (q.size() > 0) {
TreeNode node = q.poll();
TreeNode tmp = node.left;
node.left = node.right;
node.right = tmp;
if (node.left != null) {
q.add(node.left);
}
if (node.right != null) {
q.add(node.right);
}
}
return root;
}
}
``````