[LeetCode] 112. Path Sum_Easy tag: DFS

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.

典型的DFS, 然后需要将目前的path sum一起append进入stack里面, 然后判断path sum 是否跟给的sum相等并且是leaf, 即可.

1. Constraints

1) can be empty

2. IDeas

DFS:     T:O(n)    S: T(n)

3. Code

3.1) iterable

 # iterable

 class Solution:
     def pathSum(self, root, s):
         if not root: return False
         stack =[(root, root.val)]
         while stack:
             node, ps = stack.pop()
             if ps == s and not node.left and not node.right:
                 return True
             if node.left:
                 stack.append((node.left, ps + node.left.val))
             if node.right:
                 stack.append((node.right, ps + node.right.val))
         return False

3.2) recursive way.

 class Solution:
     def pathSum(self, root, s):
         if not root: return False
         if s- root.val == 0 and not root.left and not root.right:
             return True
         return self.pathSum(root.left, s- root.val) or self.pathSum(root.right, s- root.val)

4. Test cases

1) edge case

2)   s= 9

      5
     / \
    4   8