- Path Sum Easy
947
309
Favorite
Share Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
Note: A leaf is a node with no children.
Example:
Given the below binary tree and sum = 22,
5
/ \
4 8
复制代码
/ /
11 13 4 / \
7 2 1 return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
思路:深度优先算法,递减,只要中途target==0,且无左右子树,即有
代码:python3
class Solution:
def hasPathSum(self, root: TreeNode, sum: int) -> bool:
cache=[False]
def dfs(root,target):
if not root:return
target = target-root.val
if target==0 and not root.left and not root.right:
cache[0]=True
dfs(root.left,target)
dfs(root.right,target)
dfs(root,sum)
return cache[0]
复制代码
转载于:https://juejin.im/post/5d01ec5f518825276a2865f7