[LeetCode] Symmetric Tree

Problem:

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following is not:

    1
   / \
  2   2
   \   \
   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

Solution1:O(n)

public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root==null) 
           return true;
        return isSymmetric(root.left,root.right);
    }
    
    public boolean isSymmetric(TreeNode a, TreeNode b){
        if(a==null) 
           return b==null;
        if(b==null) 
           return false;
        
        if(a.val!=b.val) 
           return false;
 
        return isSymmetric(a.left,b.right)&&isSymmetric(a.right,b.left);
    }
}
Solution2:O(n)

public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root==null) 
           return true;
        Queue<TreeNode> ql = new LinkedList<TreeNode>();
        Queue<TreeNode> qr = new LinkedList<TreeNode>();
        ql.add(root.left);
        qr.add(root.right);
        while(!ql.isEmpty() && !qr.isEmpty()){
            TreeNode tmp1=ql.poll();
            TreeNode tmp2=qr.poll();
            if(tmp1==null &&tmp2!=null || tmp1!=null &&tmp2==null )
                return false;
            if(tmp1 == null)
                continue;
            if(tmp1.val!=tmp2.val) 
              return false;
            ql.add(tmp1.left);
            ql.add(tmp1.right);
            qr.add(tmp2.right);
            qr.add(tmp2.left);
        }
        
        return ql.isEmpty() && qr.isEmpty();
    }
}

Key Concept:Use Queue or traverse left and right