【Leetcode】【python】Symmetric Tree
2017-07-19 02:04
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题目大意
判断一个树是否左右对称解题思路
非递归解法
按层遍历,每一层检查一下是否对称。递归解法
其中左子树和右子树对称的条件:两个节点值相等,或者都为空
左节点的左子树和右节点的右子树对称
左节点的右子树和右节点的左子树对称
代码
递归(清晰易懂版)
class Solution(object): def judge(self, p, q): if p == None and q == None: return True if p and q and p.val == q.val: # 如果一边有一边没有那么直接返回False return self.judge(p.left, q.right) and self.judge(p.right, q.left) return False def isSymmetric(self, root): """ :type root: TreeNode :rtype: bool """ if root: return self.judge(root.left, root.right) return True
递归(超简洁版)
运算速度慢class Solution(object): def isSymmetric(self, root): """ :type root: TreeNode :rtype: bool """ def isSym(L,R): if L and R and L.val == R.val: return isSym(L.left, R.right) and isSym(L.right, R.left) return L == R return isSym(root, root)
迭代
换成数组进行判断,运算速度只超过0.5%的人class Solution(object): def isSymmetric(self, root): queue = [root] while queue: values = [i.val if i else None for i in queue] if values != values[::-1]: return False queue = [child for i in queue if i for child in (i.left, i.right)] return True
总结
递归与迭代的关系
1) 递归中一定有迭代,但是迭代中不一定有递归,大部分可以相互转换。2) 能用迭代的不用递归,递归调用函数,浪费空间,并且递归太深容易造成堆栈的溢出.函数嵌套
python允许函数嵌套,也就是说我们可以在函数内部定义一个函数,这些函数都遵循各自的作用域和生命周期规则。相关文章推荐
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