https://leetcode.com/problems/path-sum-ii/
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.
For example:
Given the below binary tree and sum = 22
,
5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2
which sum is 22.
Code
# 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 hasPathSum(self, root, sum): """ :type root: TreeNode :type sum: intyy :rtype: List[List[int]] """ if root is None: return False if root.left is None and root.right is None: return sum==root.val left = self.hasPathSum(root.left, sum-root.val) if root.left else False right = self.hasPathSum(root.right, sum-root.val) if root.right else False return left or right
Idea
Use DFS to traverse nodes.