103. Binary Tree Zigzag Level Order Traversal Leetcode Python-优快云博客

本文链接：https://blog.youkuaiyun.com/hyperbolechi/article/details/42771209

本博客详细介绍了如何实现二叉树的锯齿形层次遍历，通过层次遍历算法将所有节点值存储，并根据层级奇偶性进行反转处理。

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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]
]

假设有第0层。

这题和level order traversal的区别在于奇数level的顺序是相反的，因此做法是先用level order traversal把所有结果都存起来。然后奇数level的解求reverse.时间复杂度和空间复杂度都是O(n)

This problem is similar with level order traversal. We need to append all the value to the solution and then reverse the value based on odd levels.

Solution is following:

# 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,solution,level):
        if root:
            if len(solution)<level+1:
                solution.append([])
            solution[level].append(root.val)
            if root.left:
                self.preorder(root.left,solution,level+1)
            if root.right:
                self.preorder(root.right,solution,level+1)
    def zigzagLevelOrder(self, root):
        if root==None:
            return []
        solution=[]
        self.preorder(root,solution,0)
        for index in range(len(solution)):
            if index%2==1:
                solution[index].reverse()
        return solution