Leetcode 226) Invert Binary Tree

02 Jul 2021 in Algorithms

내 답안

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root==null){return root;}
        invert(root);
        return root;
    }
    
    public void invert(TreeNode tm){
        if(tm==null){return;}
        invert(tm.left);
        invert(tm.right);
        TreeNode tmp= tm.left;
        tm.left=tm.right;
        tm.right=tmp;
    }
}

다른 답안

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