• [leetcode]Binary Tree Zigzag Level Order Traversal @ Python


    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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  • 原文地址:https://www.cnblogs.com/zuoyuan/p/3722022.html
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