112-Path Sum

100-Same Tree | Links:

题意

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. return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

思路

• 双pre：两个先序遍历，同时左右开工的走
• 每走一层，sum减去root.val

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if not root: return False
        if not root.left and not root.right: # 走到头，结束条件
            return sum == root.val # 每走一层，sum-当前root值
        
        return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)
   

分析:

• Time: O(n)
• Space: O(n)