107. Binary Tree Level Order Traversal II(二叉树分层排序)

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Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).

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

return its bottom-up level order traversal as:

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

解法

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

class Solution:
    def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
        
        
        if not root:
            return []
        
        _result=collections.defaultdict(list)
        
        #广度遍历 BFS  
        
        def _bfs(root,level,_result):
            
  
            queue = [[root,1]]
            while len(queue) > 0:
              
                temp,level = queue.pop(0)
                _result[level].append(temp.val)
                if temp.left:
                    queue.append([temp.left,level+1])
                if temp.right:
                    queue.append([temp.right,level+1])
        _bfs(root,1,_result)
        result=[]
        for x in sorted(_result.keys(),reverse=True):
            result.append(_result[x])
        return result

上面的代码主要是通过队列实现BFS的访问，使用defaultdict对象保存结果，字典的key是对应每一层的编号，value就是每一层从左到右的排序结果。

下面给出的leetcode讨论区给出的答案，这个完全会更加易懂，实现思想都是基于BFS

class Solution(object):
    def levelOrderBottom(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        res = []
        arr = []
        if root is None:
            return []
        else:
            arr.append(root)
        while True:
            tmp = []
            while len(arr) > 0:
                item = arr.pop(0)
                tmp.append(item)
            res.insert(0, [w.val for w in tmp])
            for item in tmp:
                if item.left is not None:
                    arr.append(item.left)
                if item.right is not None:
                    arr.append(item.right)
            if len(arr) == 0:
                break
        return res

发表至：leetcode

2019-03-25

# leetcode