[leetcode]Binary Tree Zigzag Level Order Traversal @ Python

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

題意：

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}".

解題思路：這道題和層序遍歷那道題差不多，區別只是在於奇數層的節點要翻轉過來存入數組。http://www.cnblogs.com/zuoyuan/p/3722004.html

代碼：

# 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