107. Binary Tree Level Order Traversal II (Easy)

Description:

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]
]

---

Solution: 四种方案对比发现，value.insert(0,element)比value=[cache]+value要快！

class Solution:
    def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
        if root == None:
            return []
        
        value = []
        node_ls = [root]
        
        def helper(node_ls):
            cache = []
            for n in node_ls:
                if n.left != None:
                    cache.append(n.left)
                if n.right != None:
                    cache.append(n.right)
            return cache
        
        while node_ls:
            value.append([n.val for n in node_ls])
            node_ls = helper(node_ls)
            
        return value[::-1]

Runtime: 44 ms, faster than 58.17% of Python3 online submissions for Binary Tree Level Order Traversal II.
Memory Usage: 13.3 MB, less than 77.88% of Python3 online submissions for Binary Tree Level Order Traversal II.

class Solution:
    def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
        if root == None:
            return []
        
        value = []
        node_ls = [root]
        
        def helper(node_ls):
            cache = []
            for n in node_ls:
                if n.left != None:
                    cache.append(n.left)
                if n.right != None:
                    cache.append(n.right)
            return cache
        
        while node_ls:
            value.insert(0,[n.val for n in node_ls])
            node_ls = helper(node_ls)
            
        return value

Runtime: 32 ms, faster than 99.38% of Python3 online submissions for Binary Tree Level Order Traversal II.
Memory Usage: 13.4 MB, less than 72.12% of Python3 online submissions for Binary Tree Level Order Traversal II.

class Solution:
    def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
        if root == None:
            return []
        
        value = []
        node_ls = [root]
        last = root
        cache = []
        
        while node_ls:
            curr = node_ls.pop(0)
            cache.append(curr.val)
            
            if curr.left != None:
                node_ls.append(curr.left)
            if curr.right != None:
                node_ls.append(curr.right)

            if curr == last:
                value.insert(0,cache)
                cache = []
                if node_ls:
                    last = node_ls[-1]
            
        return value

Runtime: 40 ms, faster than 82.81% of Python3 online submissions for Binary Tree Level Order Traversal II.
Memory Usage: 13.6 MB, less than 25.46% of Python3 online submissions for Binary Tree Level Order Traversal II.

class Solution:
    def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
        if root == None:
            return []
        
        value = []
        node_ls = [root]
        last = root
        cache = []
        
        while node_ls:
            curr = node_ls.pop(0)
            cache.append(curr.val)
            
            if curr.left != None:
                node_ls.append(curr.left)
            if curr.right != None:
                node_ls.append(curr.right)

            if curr == last:
                value = [cache] + value
                cache = []
                if node_ls:
                    last = node_ls[-1]
            
        return value

Runtime: 44 ms, faster than 58.17% of Python3 online submissions for Binary Tree Level Order Traversal II.
Memory Usage: 13.7 MB, less than 22.73% of Python3 online submissions for Binary Tree Level Order Traversal II.