[leetcode tree]107. Binary Tree Level Order Traversal II

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],

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]
方法1：bfs

 class Solution(object):
     def levelOrderBottom(self, root):
         res,level,l= [],[[root]],[root]
         if not root:
             return res
         while level and l:
             l = [(node.left,node.right) for node in l]
             l = [v for lr in l for v in lr if v]
             if l:
                 level.append(l)
         while level:
             l = level.pop()
             r = [node.val for node in l]
             res.append(r)
         return res

方法2：bfs

 class Solution(object):
     def levelOrderBottom(self, root):
         res,level = [],[root]
         while root and level:
             res.append([node.val for node in level])
             lr = [(node.left,node.right) for node in level]
             level = [node for p in lr for node in p if node]
         res = [node for node in res[::-1]]
         return res

方法3：dfs，递归

 class Solution(object):
     def levelOrderBottom(self, root):
         res = []
         self.dfs(root,0,res)
         return res
 
     def dfs(self,root,level,res):
         if root:
             if len(res)<level+1:
                 res.insert(0,[])
             res[-(level+1)].append(root.val)
             self.dfs(root.left,level+1,res)
             self.dfs(root.right,level+1,res)

方法4：dfs,非递归

 class Solution(object):
     def levelOrderBottom(self, root):
         res = []
         stack = [(0,root)]
         while stack:
             level,node = stack.pop()
             if node:
                 if len(res) < level+1:
                     res.insert(0,[])
                 res[-(level+1)].append(node.val)
                 stack.append((level+1,node.right))
                 stack.append((level+1,node.left))
         return res