LeetCode#112路径总和 Python

最新推荐文章于 2025-12-24 20:32:08 发布

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本文探讨了在二叉树中寻找是否存在一条从根节点到叶子节点的路径，该路径上所有节点值之和等于给定的目标和。通过递归和迭代两种方法详细解析了算法实现，为理解二叉树的遍历提供了深入的视角。

在这里插入图片描述
1、递归

#class TreeNode:
#	def __init__(self, x):
#		self.val = x
#		self.left = None
#		self.right = None
class Solution:
	def hasPathSum(self, root: TreeNode, sum: int) -> bool:
		if not root: return False
		sum -= root.val
		if not root.left and not root.right:
			return sum == 0
		return self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)

2、迭代

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
        if not root: return False
        
        stack = [(root, sum - root.val)]
        while stack:
            cur_node, cur_sum = stack.pop()
            if not cur_node.left and not cur_node.right and cur_sum == 0:
                return True
            if cur_node.left:
                stack.append((cur_node.left, cur_sum - cur_node.left.val))
            if cur_node.right:
                stack.append((cur_node.right, cur_sum - cur_node.right.val))
        return False