- 分类：Tree
- 时间复杂度: O(n) 这种把树的节点都遍历一遍的情况时间复杂度为O(n)
- 空间复杂度: O(h) 树的节点的深度
112. Path Sum
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
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
代码：
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def hasPathSum(self, root: 'TreeNode', sum: 'int') -> 'bool':
if root==None:
return False
return self.helper(root,sum)
def helper(self,root,sum):
res=sum-root.val
if root.left==None and root.right==None:
if res==0:
return True
else:
return False
elif root.left==None:
return self.helper(root.right,res)
elif root.right==None:
return self.helper(root.left,res)
else:
return self.helper(root.left,res) or self.helper(root.right,res)
讨论：
1.虽然折腾了半天，但是这个代码是自己写的还是非常自豪滴！
2.出口要设置的对
