# Invert Binary Tree

ID: 226; easy

## Solution 1 (Go)

``````/**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func invertTree(root *TreeNode) *TreeNode {
if root == nil {
return nil
}
invertTree(root.Left)
invertTree(root.Right)
root.Left, root.Right = root.Right, root.Left
return root
}``````

## Solution 2 (Java)

``````/**
* 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) {
if (root == null) return;

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

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

• Recursion

## Solution 3 (Java)

``````/**
* 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) {
if (root == null) return;
Queue<TreeNode> q = new ArrayDeque<TreeNode>();
q.offer(root);

while (!q.isEmpty()) {
TreeNode curr = q.poll();
TreeNode temp = curr.left;
curr.left = curr.right;
curr.right = temp;

if (curr.left != null)
q.offer(curr.left);
if (curr.right != null)
q.offer(curr.right);
}
}
}``````

• BFS

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