100. 相同的树

广度优先遍历，依次访问2颗树的每一个节点并经行比较

 时间O(n)，空间O(n)

    public boolean isSameTree(TreeNode p, TreeNode q) {
        Deque<TreeNode> queue1 = new LinkedList<TreeNode>();
        Deque<TreeNode> queue2 = new LinkedList<TreeNode>();
        queue1.add(p);
        queue2.add(q);
        while(!queue1.isEmpty() && !queue2.isEmpty()){
            TreeNode temp1 = queue1.poll();
            TreeNode temp2 = queue2.poll();
            if (temp1==null && temp2==null) continue;
            if (temp1==null || temp2==null) return false;
            if (temp1.val!=temp2.val) return false;
            queue1.add(temp1.left);
            queue1.add(temp1.right);
            queue2.add(temp2.left);
            queue2.add(temp2.right);
        }
    return queue1.isEmpty() && queue2.isEmpty();
    }

标签：相同,queue1,queue2,add,temp2,temp1,isEmpty,100
来源： https://www.cnblogs.com/jchen104/p/14668136.html