[LeetCode] 101. Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
    1
   / \
  2   2
 / \ / \
3  4 4  3
But the following [1,2,2,null,3,null,3] is not:
    1
   / \
  2   2
   \   \
   3    3
Note:
Bonus points if you could solve it both recursively and iteratively.

Thought process:
  1. Recursive: traverse the tree and check if left matches right.
  2. Iterative: BFS.

Solution 1 (Recursive):
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        return isSymmetric(root.left, root.right);
    }
    
    private boolean isSymmetric(TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        }
        if (left == null || right == null) {
            return false;
        }
        if (left.val != right.val) {
            return false;
        }
        return isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
    }
}

Solution 2 (BFS):

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root.left);
        queue.offer(root.right);
        
        while (!queue.isEmpty()) {
            TreeNode n1 = queue.poll();
            TreeNode n2 = queue.poll();
            if (n1 == null && n2 == null) {
                continue;
            }
            if (n1 == null || n2 == null) {
                return false;
            }
            if (n1.val != n2.val) {
                return false;
            }
            
            queue.offer(n1.left);
            queue.offer(n2.right);
            queue.offer(n1.right);
            queue.offer(n2.left);
        }
        return true;
    }
}

Time complexity: O(n).
Location: Philadelphia, PA, USA

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