102. Binary Tree Level order traversal [easy] (Python) __python

Topic link

The original title of https://leetcode.com/problems/binary-tree-level-order-traversal/

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

For example:
Given binary Tree {3,9,20,#,#,15,7},

    3
   /
  9
    /   7

Return the order traversal as:

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

Title Translation

Given a binary tree, the sequence traversal results of all node values are returned. (from left to right, traversing from top to bottom)
For example, give a two-fork tree: {3,9,20,#,#,15,7},

    3
   /
  9
    /   7

Its sequence traversal results are:

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

method of Thinking idea of a

The general idea of sequence traversal is to do breadth-first traversal (BFS). Traversing the same time, note that the record traversal results to meet the requirements of the output format of the problem.

Code

# 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 to node in Q])
            new_q = [] for
            node in Q:
                if node.left:< C24/>new_q.append (node.left)
                if node.right:
                    new_q.append (node.right)
            q = new_q return

        res

Idea two

With depth-first search (DFS), the depth of a node corresponds to the subscript of an array of output results. Note that you want to save the value of each access node at the time of recursion.

Code

# 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)

PS: Novice Brush Leetcode, new handwritten blog, write wrong or write not clear also please help point out, thank you.
reprint Please specify: http://blog.csdn.net/coder_orz/article/details/51363095