二叉树的层级遍历转换

定义二叉树

class TreeNode():
    def __init__(self, val = None, left = None, right = None):
        self.left = left
        self.right = right
        self.val = val

 

1. 根据提供的列表创建成对应的二叉树，-1代表节点为空

例如:

输入:[3,9,20,-1,-1,15,7,-1,-1,-1,-1]

输出:

    3

    / \

   9  20

     /  \

    15   7

#创建二叉树
class Solution:
    def createtree(self, nums):
        if not nums: return None 
        
        new_tree = TreeNode(nums.pop(0))
        tree, left = new_tree, True
        stack = [tree]

        while stack:
            node = stack.pop(0)

            while nums:
                pop_num = nums.pop(0)

                if left:
                    if pop_num != -1:
                        node.left = TreeNode(pop_num)
                        stack.append(node.left)
                    left = False
                else:
                    if pop_num != -1:
                        node.right = TreeNode(pop_num)
                        stack.append(node.right)
                    left = True 
                
                if left:
                    break 
        
        return new_tree 
tree = Solution().createtree([3,9,20,-1,-1,15,7,-1,-1,-1,-1])

 

2. 根据二叉树层次遍历

例如:

输入:

    3

    / \

   9  20

     /  \

    15   7

输出:[3,9,20,-1,-1,15,7,-1,-1,-1,-1]

#层级遍历
class Solution:
    def leveltravel(self, tree):
        if not tree: return []

        results = [tree.val]
        stack = [tree]

        while stack:
            node = stack.pop(0)

            if node.left: 
                results.append(node.left.val)
                stack.append(node.left)
            else:
                results.append(-1) 
                
            if node.right:
                results.append(node.right.val)
                stack.append(node.right)
            else:
                results.append(-1) 

        return results           

# tree = TreeNode(3, TreeNode(9), TreeNode(20, TreeNode(15), TreeNode(7)))
res = Solution().leveltravel(tree)
print(res)