path-sum.py

# -*- coding:utf-8 -*-


# 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.
#


# 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: int
        :rtype: bool
        """
        if not root:
            return False

        stack = [(root, sum)]
        while stack:
            node, sum = stack.pop()
            if not node.left and not node.right and node.val == sum:
                return True
            
            if node.left:
                stack.append((node.left, sum-node.val))
            if node.right:
                stack.append((node.right, sum-node.val))
            
        return False