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.
思路: 递归的暴力破解。
穷尽式的解决了所有子问题后,就放心的自我调用吧。
# Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: # @param root, a tree node # @param sum, an integer # @return a boolean def hasPathSum(self, root, sum): if root == None: return False if root.val == sum and root.left == None and root.right == None: return True return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)