题解 | #二叉树中和为某一值的路径(一)#

利用广度优先遍历将节点的值压入栈，当遇到相等的值时弹栈，最后栈中午元素说明对称（注意空节点处理）

#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param pRoot TreeNode类 
# @return bool布尔型
#
class Solution:
    def isSymmetrical(self , pRoot: TreeNode) -> bool:
        nodeQueue=[pRoot]
        valStack=[]
        if not pRoot:
            return True
        while len(nodeQueue)>0:
            if nodeQueue[0].left:
                nodeQueue.append(nodeQueue[0].left)
                if len(valStack)>0 and nodeQueue[0].left.val==valStack[-1]:
                    valStack.pop()
                else:
                    valStack.append(nodeQueue[0].left.val)
            else:
                if  len(valStack)>0 and valStack[-1]=='#':
                    valStack.pop(-1)
                else:
                    valStack.append('#')
            if nodeQueue[0].right:
                nodeQueue.append(nodeQueue[0].right)
                if len(valStack)>0 and nodeQueue[0].right.val==valStack[-1]:
                    valStack.pop()
                else:
                    valStack.append(nodeQueue[0].right.val)
            else:
                if len(valStack)>0 and valStack[-1]=='#':
                    valStack.pop(-1)
                else:
                    valStack.append('#')
            nodeQueue.pop(0)
        if len(valStack)>0:
            return False
        else:
            return True
        # write code here