LeetCode 100. Same Tree

Given two binary trees, write a function to check if they are equal or not.

Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

public class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        //非循环实现
        /*if(p==null && q==null){  //两棵空树
            return true;
        }
        else if(p==null || q==null){  //一棵空树，一棵非空树
            return false;
        }
        else{               //两棵非空树
            Stack<TreeNode> s1=new Stack<TreeNode>();
            Stack<TreeNode> s2=new Stack<TreeNode>();
            s1.push(p);
            s2.push(q);
            try{
                while(s1!=null && s2!=null){
                TreeNode ps=s1.pop();
                TreeNode qs=s2.pop();
                if(ps.val!=qs.val){
                    return false;
                }else{
                    if(ps.left!=null && qs.left!=null){ //左子树同时不为空
                        s1.push(ps.left);
                        s2.push(qs.left);
                    }else if(ps.left!=null || qs.left!=null){ //左子树不同时为空
                        return false;
                    }
                    if(ps.right!=null && qs.right!=null){ //右子树同时为空
                        s1.push(ps.right);
                        s2.push(qs.right);
                    }else if(ps.right!=null || qs.right!=null){ //右子树不同时为空
                        return false;
                    }
                }
            }
                if(s1==null && s2==null){  //两棵树的节点一样多
                return true;
                 }else{
                return false;
                 }
            
            }catch(Exception e){
                
            }
            
            return true;
        }*/
        
        //循环实现
        if(p==null && q==null){
            return true;
        }
        else if(p==null || q==null){
            return false;
        }
        else{
            if(p.val==q.val){
            return (isSameTree(p.left,q.left)&& isSameTree(p.right,q.right));
            }else{
                return false;
            }
            
        }
    }
}