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.node
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
注意这里的路径必须是从根到叶子(必须到叶子!)。这里有个神奇的现象sum -= root.val
这句不谢的话,下面改为if sum == root.val and root.left == None and root.right == None:
就过不了it
# 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): ret = False if root == None: return ret sum -= root.val if sum == 0 and root.left == None and root.right == None: ret = True return ret or self.hasPathSum(root.left,sum) or self.hasPathSum(root.right,sum)