102. Binary Tree Level Order Traversal [easy] (Python)__Python

題目連結

https://leetcode.com/problems/binary-tree-level-order-traversal/ 題目原文

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,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its level order traversal as:

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

題目翻譯

給定一個二叉樹，返回所有節點值的層序遍曆結果。（從左至右，從上到下遍曆）
比如，給出二叉樹：{3,9,20,#,#,15,7}，

    3   / \  9  20    /  \   15   7

它的層序遍曆結果為：

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

思路方法 思路一

層序遍曆的一般想法，做廣度優先遍曆（BFS）。遍曆的同時，注意記錄遍曆結果要用滿足題目要求的輸出格式。

代碼

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

思路二

用深度優先搜尋（DFS），節點的深度與輸出結果數組的下標相對應。注意在遞迴的時候要儲存每次訪問的節點值。

代碼

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass 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: 新手刷LeetCode，新手寫部落格，寫錯了或者寫的不清楚還請幫忙指出，謝謝。
轉載請註明：http://blog.csdn.net/coder_orz/article/details/51363095