leetcode 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]
]

# 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 levelOrderBottom(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root: 
            return []
        q = [root]
        ans = []
        while q:
            ans.append([n.val for n in q])
            q = [n for node in q for n in (node.left, node.right) if n]           
        return ans[::-1]

上面解法是层序遍历，使用先序遍历，递归：

# 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 levelOrderBottom(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """        
        def dfs(node, paths, depth):
            if not node: return
            if len(paths) == depth:
                paths.append([])                
            paths[depth].append(node.val)
            dfs(node.left, paths, depth+1)
            dfs(node.right, paths, depth+1)
        ans = []
        dfs(root, ans, 0)
        return ans[::-1]