Leetcode 103.二叉树的锯齿形层次遍历

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103.二叉树的锯齿形层次遍历

题目

给定一个二叉树，返回其节点值的锯齿形层次遍历。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。

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例如：
给定二叉树 [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7
返回锯齿形层次遍历如下：

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

方法

方法1：宽度优先遍历 & dict

Using dic to store the values of each level’s nodes in corresponding lists. When return the result, reverse every other level.

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# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def zigzagLevelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        queue=collections.deque([(root,0)]) if root else []
        dic=collections.defaultdict(list)
        while queue:
            node,level=queue.popleft()
            dic[level].append(node.val)
            if node.left:
                queue.append((node.left,level+1))
            if node.right:
                queue.append((node.right,level+1))
        return [dic[level][::(1,-1)[level&1]] for level in range(len(dic))]

方法2：宽度优先遍历 & list

Simple straightforward solution using flag to decide whether from left to right or from right to left

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# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def zigzagLevelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []

        res,tmp,queue,flag=[],[],collections.deque([root]),1
        while queue:
            for i in range(len(queue)):
                node=queue.popleft()
                tmp+=[node.val]
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            res+=[tmp[::flag]]
            tmp=[]
            flag*=-1
        return res
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class Solution(object):
def zigzagLevelOrder(self, root):
    """
    :type root: TreeNode
    :rtype: List[List[int]]
    """
    if not root: return []
    res, temp, stack, flag=[], [], [root], 1
    while stack:
        for i in xrange(len(stack)):
            node=stack.pop(0)
            temp+=[node.val]
            if node.left: stack+=[node.left]
            if node.right: stack+=[node.right]
        res+=[temp[::flag]]
        temp=[]
        flag*=-1
    return res
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# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def zigzagLevelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        _r = []
        if not root:
            return _r

        _q = []
        _q.append(root)

        while _q:
            tmp = []

            for i in range(len(_q)):
                _c = _q.pop(0)
                tmp.append(_c.val)

                if _c.left:
                    _q.append(_c.left)
                if _c.right:
                    _q.append(_c.right)

            _r.append(tmp)

        for index, elem in enumerate(_r):
            if index%2:
                _r[index].reverse()
        return _r