102..103.199二叉树的(锯齿形）层次遍历、右视图（中等，树）

102、给定一个二叉树，返回其按层次遍历的节点值（即逐层地，从左到右访问所有节点）。

例如:
给定二叉树: [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

返回其层次遍历结果：

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

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

class Solution:
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        result=[]
        currentnode=[]
        if not root:
            return result
        currentnode.append(root)
        while currentnode:
            single=[]
            l=len(currentnode)
            for i in range(l):
                current=currentnode.pop(0)
                if current.left:
                    currentnode.append(current.left)
                if current.right:
                    currentnode.append(current.right)
                single.append(current.val)
            result.append(single)
        return result

103、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）。

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]
]

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

class Solution:
    def zigzagLevelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        result=[]
        currentnode=[]
        if not root:
            return result
        currentnode.append(root)
        j=-1
        while currentnode:
            single=[]
            l=len(currentnode)
            for i in range(l):
                current=currentnode.pop(0)
                if current.left:
                    currentnode.append(current.left)
                if current.right:
                    currentnode.append(current.right)
                single.append(current.val)
            if j>0:
                single=single[::-1]
            j=-j
            result.append(single)
        return result

199、给定一棵二叉树，想象自己站在它的右侧，按照从顶部到底部的顺序，返回从右侧所能看到的节点值。

示例:

输入: [1,2,3,null,5,null,4]
输出: [1, 3, 4]
解释:

   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---

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

class Solution:
    def rightSideView(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        result=[]
        currentnode=[root]
        while currentnode:
            single=[]
            l=len(currentnode)
            for i in range(l):
                current=currentnode.pop(0)
                if current.left:
                    currentnode.append(current.left)
                if current.right:
                    currentnode.append(current.right)
                single.append(current.val)
            result.append(single)
        return [s[-1] for s in result]

执行用时: 56 ms, 在Binary Tree Right Side View的Python3提交中击败了71.15%的用户

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

class Solution:
    def rightSideView(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        ret = [root.val]
        left = self.rightSideView(root.left)
        right = self.rightSideView(root.right)
        ret  += right + left[len(right):]          
        return ret

第二个方法很简单