lintcode-easy-Invert Binary Tree

Invert a binary tree.

Example

  1         1
 / \       / 2   3  => 3   2
   /         4         4

Challenge

Do it in recursion is acceptable, can you do it without recursion?

1. 递归

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: a TreeNode, the root of the binary tree
     * @return: nothing
     */
    public void invertBinaryTree(TreeNode root) {
        // write your code here
        if(root == null)
            return;

        invertBinaryTree(root.left);
        invertBinaryTree(root.right);

        TreeNode temp = root.right;

        root.right = root.left;
        root.left = temp;

        return;
    }
}

2. 非递归

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: a TreeNode, the root of the binary tree
     * @return: nothing
     */
    public void invertBinaryTree(TreeNode root) {
        // write your code here
        if(root == null)
            return;

        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);

        while(!queue.isEmpty()){
            TreeNode p = queue.poll();

            if(p.left != null)
                queue.offer(p.left);
            if(p.right != null)
                queue.offer(p.right);

            TreeNode temp = p.right;
            p.right = p.left;
            p.left = temp;
        }

        return;
    }
}