226.Invert Binary Tree

Solution 1:DFS, Recursion On

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root == null){
            return null;
        }
        else{
            TreeNode newLeft = invertTree(root.right);
            TreeNode newRight = invertTree(root.left);
            root.left = newLeft;
            root.right = newRight;
            return root;
        }
    }
}

Solution 2 BFS On

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root == null){
            return null;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        while(!queue.isEmpty()){
            TreeNode current = queue.poll();
            TreeNode temp = current.left;
            current.left = current.right;
            current.right = temp;
            if(current.left != null){
                queue.add(current.left);
            }
            if(current.right != null){
                queue.add(current.right);
            }
        }
        return root;
    }
}