LeetCode Trees and Graphs: Problem #101: Symmetric Tree

The next problem on hit list is Problem #101: Symmetric Tree

Again, this can be solved using many approaches both recursively or iteratively.

Approach 1: My recursive solution with complexity O(n) where n is the total number of TreeNodes in the tree is as follows:

/**
 * 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 boolean isSymmetric(TreeNode root) {
        if (root == null)
            return true;
        
        if (root.left == null && root.right == null)
            return true;
        
        return isMirror (root.left, root.right);
    }
    
    private boolean isMirror(TreeNode r1, TreeNode r2) {
        if (r1 == null && r2 == null)
            return true;
        if ((r1 == null && r2 != null) || (r1 != null && r2 == null))
            return false;
        if (r1.val != r2.val)
            return false;
        
        boolean mirror = isMirror(r1.left, r2.right);
        if (mirror)
            mirror = isMirror(r1.right, r2.left);
        
        return mirror;
    }
}

Approach 2: An iterative solution approach making use of a BFS like queue insertion of node-pairs to check for equality when popped (again O(n)):

/**
 * 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 boolean isSymmetric(TreeNode root) {
        if (root == null)
            return true;
        
        if (root.left == null && root.right == null)
            return true;
        
        //iterative solution
        
        Queue<TreeNode> q = new LinkedList<>();
        q.add(root.left);
        q.add(root.right);
        while(!q.isEmpty()) {
            TreeNode n1 = q.remove();
            TreeNode n2 = q.remove();
            
            if (n1 == null && n2 == null)
                continue;
            if ((n1 == null && n2 != null) || (n1 != null && n2 == null))
                return false;
            if (n1.val != n2.val)
                return false;
            
            q.add(n1.left);
            q.add(n2.right);
            q.add(n1.right);
            q.add(n2.left);
        }
        
        return true;
    }
}

That's not all! There are definitely more approaches to it. One which I can think of is using stacks. Feel free to share your solutions for the same.

As always, you can checkout my latest accepted solutions on leetcode at https://leetcode.com/chandniverma/ . All the best!