递归与二叉树_leetcode101

最新推荐文章于 2025-09-06 22:33:46 发布

weixin_30653097

最新推荐文章于 2025-09-06 22:33:46 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/lux-ace/p/10546763.html

此博客内容为转载，原文链接为https://www.cnblogs.com/lux-ace/p/10546763.html ，标签涉及数据结构与算法。

#coding=utf-8
# Definition for a binary tree node.
class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
        self.right = None


# 递归
class Solution1(object):
def isSymmetric(self, root):
"""
        :type root: TreeNode
        :rtype: bool
        """


        return self.isSame(root,root)

# coding=utf-8
    # Definition for a binary tree node.
    class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
            self.right = None

    # 递归
    class Solution1(object):
def isSymmetric(self, root):
"""
            :type root: TreeNode
            :rtype: bool
            """

            return self.isSame(root, root)

def isSame(self, left, right):

if not left and not right:
return True
            if not left and right:
return False
            if left and not right:
return False
            if left.val == right.val:
return self.isSame(left.left, right.right) and self.isSame(left.right, right.left)
else:
return False

    # 迭代
    class Solution2(object):
def isSymmetric(self, root):
"""
            :type root: TreeNode
            :rtype: bool
            """
            if not root:
return True
            else:
return self.isMirror(root.left, root.right)

def isMirror(self, left, right):
            queue = []
            queue.append(left)
            queue.append(right)

# 每次成对的比较
            while queue:

                curLeft = queue.pop(0)
                curRight = queue.pop(0)

if not curLeft and not curRight:
continue
                if not curLeft or not curRight or curLeft.val != curRight.val:
return False

                queue.append(curLeft.left)
                queue.append(curRight.right)
                queue.append(curLeft.right)
                queue.append(curRight.left)

return True

        def createTree(self):

            root = TreeNode(1)
            root.left = TreeNode(2)
            root.right = TreeNode(2)

            root.left.left = TreeNode(3)
            root.left.right = TreeNode(4)
            root.right.left = TreeNode(4)
            root.right.right = TreeNode(3)

return root

def printTree(self, root):
            res = []

if not root:
return res

            queue = []
            queue.append(root)

while queue:
                level = []
for i in range(len(queue)):
                    curNode = queue.pop(0)
                    level.append(curNode.val)

if curNode.left:
                        queue.append(curNode.left)
if curNode.right:
                        queue.append(curNode.right)

                res.append(level)

return res

    s = Solution2()

    root = s.createTree()

# print s.printTree(root)

    print  s.isSymmetric(root)
def isSame(self,left,right):

if not left and not right:
return True
        if not left and right:
return False
        if left and not right:
return False
        if left.val == right.val:
return self.isSame(left.left,right.right) and self.isSame(left.right,right.left)
else:
return False



# 迭代
class Solution2(object):
def isSymmetric(self, root):
"""
        :type root: TreeNode        :rtype: bool        """        if not root:            return True        else:            return self.isMirror(root.left,root.right)    def isMirror(self,left,right):        queue = []        queue.append(left)        queue.append(right)        # 每次成对的比较        while queue:            curLeft = queue.pop(0)            curRight = queue.pop(0)            if not curLeft and not curRight:                continue            if not curLeft or not curRight or curLeft.val != curRight.val:                return False            queue.append(curLeft.left)            queue.append(curRight.right)            queue.append(curLeft.right)            queue.append(curRight.left)        return True    def createTree(self):        root = TreeNode(1)        root.left = TreeNode(2)        root.right = TreeNode(2)        root.left.left= TreeNode(3)        root.left.right = TreeNode(4)        root.right.left = TreeNode(4)        root.right.right = TreeNode(3)        return root    def printTree(self,root):        res = []        if not root:            return res        queue = []        queue.append(root)        while queue:            level = []            for i in range(len(queue)):                curNode = queue.pop(0)                level.append(curNode.val)                if curNode.left:                    queue.append(curNode.left)                if curNode.right:                    queue.append(curNode.right)            res.append(level)        return ress = Solution2()root = s.createTree()# print s.printTree(root)print  s.isSymmetric(root)