101. Symmetric Tree

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

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

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

But the following [1,2,2,null,3,null,3] is not:

    1
   / \
  2   2
   \   \
   3    3

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

解法

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        
        if not root:
            return True

        dq = collections.deque([(root.left,root.right),])
        while dq:
            node1, node2 = dq.popleft()
            #两个都是空节点直接跳过
            if not node1 and not node2:
                continue
            #只要其中一个节点为空直接返回False
            if not node1 or not node2:
                return False
            #若两个节点都存在,那么需要判断两者的值是否相等
            if node1.val != node2.val:
                return False
            # 判断左节点的左节点与右节点的右节点
            # 判断左节点的右节点和右节点的左节点
            dq.append((node1.left,node2.right))
            dq.append((node1.right,node2.left))
        return True
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