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.
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.
Solution:
1. use iterative way; bfs
1 if root is None: 2 return False 3 d = deque() 4 5 d.append((root, root.val)) 6 7 while(len(d)): 8 nodeInfo = d.popleft() 9 node = nodeInfo[0] 10 s = nodeInfo[1] 11 if (not node.left) and (not node.right): 12 if s == sum: 13 return True 14 if node.left: 15 d.append((node.left, s + node.left.val)) 16 if node.right: 17 d.append((node.right, s + node.right.val)) 18 return False
2. use recursive way
1 return self.haspathSumHelper(root, 0, sum) 2 3 def haspathSumHelper(self, node, current, sum): 4 5 if node is None: 6 return False 7 current += node.val 8 if (not node.left ) and (not node.right): 9 if current == sum: 10 return True 11 else: 12 return False 13 14 15 return self.haspathSumHelper(node.left, current, sum) or self.haspathSumHelper(node.right, current, sum)