lintcode 二叉树的层次遍历

问题描述

http://www.lintcode.com/zh-cn/problem/binary-tree-level-order-traversal/#
给出一棵二叉树，返回其节点值的层次遍历（逐层从左往右访问）
样例
给一棵二叉树 {3,9,20,#,#,15,7} ：

3
/ \
9 20
/ \
15 7
返回他的分层遍历结果：

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

先从最简单的问题开始

遍历二叉树的第k层，相当于遍历二叉树根节点的左右子树的第k-1层。这样一直遍历下去，直到k=0时，输出节点即可。

def printnode(TreeNode root, int k,list l):    if(not root or k<0):        return 0    if(k==0):        l.append([root.val])        return 0    else:         l.append([printnode(root.left,k-1), printnode(root.left,k-1)]        return 0

层次遍历

写一个for循环即可

for i in range(max_depth):    printnode(root,i,l)

完整代码

"""Definition of TreeNode:class TreeNode:    def __init__(self, val):        self.val = val        self.left, self.right = None, None"""class Solution:    """    @param: root: A Tree    @return: Level order a list of lists of integer    """    """Definition of TreeNode:class TreeNode:    def __init__(self, val):        self.val = val        self.left, self.right = None, None"""class Solution:    """    @param: root: A Tree    @return: Level order a list of lists of integer    """          def printnode(TreeNode root, int k,list l):        if(not root or k<0):            return 0        if(k==0):            l.append([root.val])            return 0        else:             l.append([printnode(root.left,k-1), printnode(root.left,k-1)]            return 0    def levelOrder(self, root):        # write your code here        max_depth=3        queue=[]        for i in range(max_depth):            self.printnode(self.root,i,queue)        return queue