Leetcode【102】Binary Tree Level Order Traversal（Python版）

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本文介绍了一种实现二叉树层次遍历的算法，通过使用栈来存储每层的节点，最后输出每一层的节点值。此算法首先将根节点放入栈中，然后不断迭代，直到栈为空。在每次迭代中，会将当前层所有节点的子节点加入到下一个栈中，同时收集当前层的节点值，最终形成层次遍历的结果。

基本思路就是：

遍历当前层的所有根节点组成的栈，形成下一层的根节点组成的栈，outList不断叠加每一层的根节点组合的列表

# 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 levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []
        currentStack = [root]
        outList= []
        while currentStack:
            nextStack = []
            tmp = []
            for point in currentStack:
                if point.left:
                    nextStack.append(point.left)
                if point.right:
                    nextStack.append(point.right)
                tmp.append(point.val)
            outList.append(tmp)
            currentStack = nextStack

        return outList

完整测试代码

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

'''
    3
   / \
  9  20
    /  \
   15   7
'''


class Solution(object):
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []
        currentStack = [root]
        outList= []
        while currentStack:
            nextStack = []
            tmp = []
            for point in currentStack:
                if point.left:
                    nextStack.append(point.left)
                if point.right:
                    nextStack.append(point.right)
                tmp.append(point.val)
            outList.append(tmp)
            currentStack = nextStack

        return outList


n1 = TreeNode(3)
n2 = TreeNode(9)
n3 = TreeNode(20)
n4 = TreeNode(15)
n5 = TreeNode(7)
n1.left = n2
n1.right = n3
n3.left = n4
n3.right = n5
s = Solution()
result = s.levelOrder(n1)
print(result)