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[leetcode]Binary Tree Zigzag Level Order Traversal @ Python

2024-07-04 02:44:35   225人阅读

原题地址：http://oj.leetcode.com/problems/binary-tree-zigzag-level-order-traversal/

题意：

Given a binary tree, return the zigzag level order traversal of its nodes‘ values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   /   9  20
    /     15   7

 

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]

 

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.


OJ‘s Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.

Here‘s an example:

   1
  /  2   3
    /
   4
         5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".
解题思路：这道题和层序遍历那道题差不多，区别只是在于奇数层的节点要翻转过来存入数组。http://www.cnblogs.com/zuoyuan/p/3722004.html
代码：
# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param root, a tree node
    # @return a list of lists of integers
    def preorder(self, root, level, res):
        if root:
            if len(res) < level+1: res.append([])
            if level % 2 == 0: res[level].append(root.val)
            else: res[level].insert(0, root.val)
            self.preorder(root.left, level+1, res)
            self.preorder(root.right, level+1, res)
    def zigzagLevelOrder(self, root):
        res=[]
        self.preorder(root, 0, res)
        return res

 


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