Leetcode 101 symmetric binary tree

recursive

var isSymmetric = function(root) {
    
    return dfs(root, root);
};

//  The condition of symmetry is that the left subtree of a tree is mirrored symmetrically with its right subtree 
//  Both pointers start from the root node at the same time 
//  One to the left , One to the right 
//  If the two values are different , Then it means it is not symmetrical 
const dfs = (p, q) => {
    
    if (p === null && q === null) return true;
    if (p === null || q === null) return false;
    return p.val === q.val && dfs(p.left, q.right) && dfs(p.right, q.left);
}

Guang Shu

var isSymmetric = function(root) {
    
    let queue = [root, root];
    while (queue.length > 0) {
    
        let l = queue.shift();
        let r = queue.shift();
        
        if (l === null && r === null) {
    
            continue;
        }
        if (l === null || r === null || l.val !== r.val) {
    
            return false;
        }

        //  Each time you join the team, first join the left node of the first pointer , And the right node of the second pointer 
        //  Make sure to compare the left node with the right node when taking out the first two of the team , This is mirror symmetric 
        queue.push(l.left);
        queue.push(r.right);
        //  Then queue the right node of the first pointer , And the left node of the second pointer 
        queue.push(l.right);
        queue.push(r.left);
    }
    return true;
};