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
1 class Solution(object): 2 def levelOrderBottom(self, root): 3 res,level,l= [],[[root]],[root] 4 if not root: 5 return res 6 while level and l: 7 l = [(node.left,node.right) for node in l] 8 l = [v for lr in l for v in lr if v] 9 if l: 10 level.append(l) 11 while level: 12 l = level.pop() 13 r = [node.val for node in l] 14 res.append(r) 15 return res
方法2：bfs
1 class Solution(object): 2 def levelOrderBottom(self, root): 3 res,level = [],[root] 4 while root and level: 5 res.append([node.val for node in level]) 6 lr = [(node.left,node.right) for node in level] 7 level = [node for p in lr for node in p if node] 8 res = [node for node in res[::-1]] 9 return res
方法3：dfs，递归
1 class Solution(object): 2 def levelOrderBottom(self, root): 3 res = [] 4 self.dfs(root,0,res) 5 return res 6 7 def dfs(self,root,level,res): 8 if root: 9 if len(res)<level+1: 10 res.insert(0,[]) 11 res[-(level+1)].append(root.val) 12 self.dfs(root.left,level+1,res) 13 self.dfs(root.right,level+1,res)
方法4：dfs,非递归
1 class Solution(object): 2 def levelOrderBottom(self, root): 3 res = [] 4 stack = [(0,root)] 5 while stack: 6 level,node = stack.pop() 7 if node: 8 if len(res) < level+1: 9 res.insert(0,[]) 10 res[-(level+1)].append(node.val) 11 stack.append((level+1,node.right)) 12 stack.append((level+1,node.left)) 13 return res