[LeetCode] Binary Tree Level Order Traversal - Python

原题：https://leetcode.com/problems/binary-tree-level-order-traversal/description/

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

即打印二叉树结构

代码：

广度优先

# 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 levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        res = []
        if root == None:
            return res

        q = [root]
        while len(q) != 0:
            res.append([node.val for node in q])
            new_q = []
            for node in q:
                if node.left:
                    new_q.append(node.left)
                if node.right:
                    new_q.append(node.right)
            q = new_q

        return res

深度优先

# 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 levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        res = []
        self.dfs(root, 0, res)
        return res

    def dfs(self, root, depth, res):
        if root == None:
            return res
        if len(res) < depth+1:
            res.append([])
        res[depth].append(root.val)
        self.dfs(root.left, depth+1, res)
        self.dfs(root.right, depth+1, res)