Thinking process

Iterative solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);

        while (!queue.isEmpty()) {
            int size = queue.size(); 

            List<Integer> level = new ArrayList<>();
            for (int i = 0; i < size; ++i) {
                TreeNode node = queue.poll();

                if (node.left != null) {
                    queue.offer(node.left);
                    level.add(node.left.val);
                } else {
                    level.add(0);
                }

                if (node.right != null) {
                    queue.offer(node.right);
                    level.add(node.right.val);
                } else {
                    level.add(0);
                }
            }

            if (!checkLevel(level)) {
                return false;
            }

        }

        return true;

    }

    private boolean checkLevel(List<Integer> list) {
        int i = 0, j = list.size() - 1;

        while (i <= j) {
            if (list.get(i) != list.get(j)) {
                return false;
            }
            i++;
            j--;
        }

        return true;
    }

}

/////////////////////////////////////2////////////////////////////////////////
public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        Queue<TreeNode> queue1 = new LinkedList<>();
        Queue<TreeNode> queue2 = new LinkedList<>();

        queue1.offer(root.left);
        queue2.offer(root.right);

        while (!queue1.isEmpty() && !queue2.isEmpty()) {
            TreeNode node1 = queue1.poll();
            TreeNode node2 = queue2.poll();

            if (node1 == null && node2 == null) continue;
            if (node1 == null || node2 == null) return false;

            if (node1.val != node2.val) return false;

            queue1.offer(node1.left);
            queue2.offer(node2.right);

            queue1.offer(node1.right);
            queue2.offer(node2.left);
        }

        return queue1.isEmpty() && queue2.isEmpty();
    }
}

Recursion

public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;

        return isSubTreeSymmetric(root.left, root.right);
    }

    private boolean isSubTreeSymmetric(TreeNode left, TreeNode right) {
        if (left == null || right == null) {
            return left == right;
        } else {
            if (left.val != right.val) return false;
        }

        return isSubTreeSymmetric(left.left, right.right) && isSubTreeSymmetric(left.right, right.left);
    }
}

////////////////////////////////////////////////2///////////////////////////////////////////
public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;

        return isSymmetricDfsHelper(root.left, root.right);
    }

    //compare node1 and node 2
    //return true, if node1.val == node2.val and node1.left.val == node2.right.val and node1.right.val == node2.left.val
    //else return false
    private boolean isSymmetricDfsHelper(TreeNode node1, TreeNode node2) {
        if (node1 == null && node2 == null) return true;
        if (node1 == null || node2 == null) return false;

        return node1.val == node2.val && isSymmetricDfsHelper(node1.left, node2.right) && isSymmetricDfsHelper(node1.right, node2.left);
    }
}

Follow up: what if there are three children of each node, left, mid and right?

• use the level order traversal solution iteratively