Invert Binary Tree

Invert a binary tree.

Have you met this question in a real interview?

Yes

Example

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

Challenge

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

/**

 * 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;

        TreeNode temp = root.left;

        root.left = root.right;

        root.right = temp;

        invertBinaryTree(root.left);

        invertBinaryTree(root.right);

    }

}

DFS

/**

 * 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;

        Stack<TreeNode> stack = new Stack<TreeNode>();

        stack.push(root);

        while(!stack.isEmpty()){

            TreeNode node = stack.pop();

            if(node == null) continue;

            TreeNode temp = node.left;

            node.left = node.right;

            node.right = temp;

            stack.push(node.left);

            stack.push(node.right);

        }

    }

}

BFS

/**

 * 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;

        Queue<TreeNode> queue = new LinkedList<TreeNode>();

        queue.offer(root);

        while(!queue.isEmpty()){

            TreeNode node = queue.poll();

            if(node == null) continue;

            TreeNode temp = node.left;

            node.left = node.right;

            node.right = temp;

            queue.offer(node.left);

            queue.offer(node.right);

        }

    }

}