topic:node
101. Symmetric TreeDescription:this
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).For examplecode
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.
解題思路:1.所謂的對稱,是左右相反位置的節點的值判斷是否相同。ip
2.全部的節點對稱,是能夠從源頭追根溯源的。 3.只要出現不一樣,便可返回便可,不然繼續進行處理。
代碼以下:it
# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None from collections import deque class Solution: def isSymmetric(self, root): """ :type root: TreeNode :rtype: bool """ if not root: return True nodes_stack=[root.left,root.right] while nodes_stack: val_left,val_right=nodes_stack.pop(0),nodes_stack.pop(0) if not val_left and not val_right: continue elif not val_left or not val_right: return False elif val_left.val!=val_right.val: return False else: nodes_stack.extend([val_left.left,val_right.right,val_left.right,val_right.left]) return True