标签：continue while container translate problems turn contain end span
给定一个二叉树和一个目标和，判断该树中是否存在根节点到叶子节点的路径，这条路径上所有节点值相加等于目标和。
说明: 叶子节点是指没有子节点的节点。
示例:
给定如下二叉树，以及目标和 sum = 22，
5
/ 4 8
/ / 11 13 4
/ \ 7 2 1
返回 true, 因为存在目标和为 22 的根节点到叶子节点的路径 5->4->11->2。
class Solution: def hasPathSum(self, root: TreeNode, sum: int) -> bool: if not root: return False if not root.left and not root.right: return sum == root.val return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)
class Solution: def hasPathSum(self, root: TreeNode, sum: int) -> bool: if not root: return False que_node = collections.deque([root]) que_val = collections.deque([root.val]) while que_node: now = que_node.popleft() temp = que_val.popleft() if not now.left and not now.right: if temp == sum: return True continue if now.left: que_node.append(now.left) que_val.append(now.left.val + temp) if now.right: que_node.append(now.right) que_val.append(now.right.val + temp) return False
标签：continue while container translate problems turn contain end span
原文地址：https://www.cnblogs.com/topass123/p/13258717.html
