LeetCode-Python-102. 二叉树的层次遍历

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]
]

思路：

BFS, 处理每一层的时候记录下一层的节点，并把当前这一层每个节点的值记录到结果里。

class Solution(object):
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []
        node = [root]
        result = list(list())
        self.generate(node, result)
        return result
    
    def generate(self, node, result):
        next_layer_node = []
        current_layer_result = []

        for node in node:
            if node.left:
                next_layer_node.append(node.left)
            if node.right:
                next_layer_node.append(node.right)
            current_layer_result.append(node.val)

        result.append(current_layer_result)
        if len(next_layer_node) == 0:
            return

        self.generate(next_layer_node, result)

下面的写于2019.6.24 

class Solution(object):
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        queue = [root]
        res = []
        while queue:
            next_queue = []
            layer = []
            for node in queue:
                if node:
                    layer.append(node.val)
                    next_queue += [node.left, node.right]
            queue = next_queue[:]
            if layer:
                res.append(layer[:])
        return res