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;
    }
}