# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
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1.用一个栈来操作两棵树
2.每次入栈同时压入两棵树的节点
3.每次去除同时取出两棵树的节点，这样和操作一棵树是一样的。
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class Solution(object):
    def isSameTree(self, p, q):
        """
        :type p: TreeNode
        :type q: TreeNode
        :rtype: bool
        """
        if p is None and q is None:return True
        if p is None and q: return False
        if p and q is None :return False
        que = []
        que.append(p)
        que.append(q)
        while que:
            nodep = que.pop(0)
            nodeq = que.pop(0)
            #当前节点一定是存在的，只需要对比值是否相同就行
            if nodep.val!=nodeq.val:return False
                        
            #先是p,q的左子节点存在->入栈
            #(不管值对不对，这里只关心两棵树的子节点是否都存在，值的正确与否交给上一行代码)
            #然后列出两棵树的子节点不一致的情况（笨办法，全列出来了。。。因为有可能遇到叶子节点）           
            if nodep.left and nodeq.left:
                que.append(nodep.left)
                que.append(nodeq.left)
            elif nodep.left and nodeq.left is None:return False
            elif nodep.left is None and nodeq.left:return False
            if nodep.right and nodeq.right:
                que.append(nodep.right)
                que.append(nodeq.right)
            elif nodep.right and nodeq.right is None:return False
            elif nodep.right is None and nodeq.right : return False
        #全部没问题，返回True
        return True