LeetCode-Path Sum I & II & III

112. Path Sum

https://leetcode.com/problems/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

# 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):
    # iteratively
    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 and root.val == sum:
            return True
        stack = [(root, root.val)]
        while stack:
            node, val = stack.pop()
            if not node.left and not node.right and val == sum:
                return True
            if node.left:
                stack.append((node.left, val + node.left.val))
            if node.right:
                stack.append((node.right, val + node.right.val))
        return False
    
    '''
    # recursively
    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 and root.val == sum:
            return True
        return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)
      ''' 

113. Path Sum II

https://leetcode.com/problems/path-sum-ii/

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

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

return

[
   [5,4,11,2],
   [5,8,4,5]
]

Solution

# 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 pathSum(self, root, s):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        if not root:
            return []
        res = []
        stack = [(root, [root.val])]
        while stack:
            cur, ls = stack.pop()
            if not cur.left and not cur.right and sum(ls) == s:
                res.append(ls)
            if cur.left:
                stack.append((cur.left, ls+[cur.left.val]))
            if cur.right:
                stack.append((cur.right, ls+[cur.right.val]))
        return res
    '''
    # recursively
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        if not root:
            return []
        if not root.left and not root.right and sum == root.val:
            return [[root.val]]
        temp = self.pathSum(root.left, sum-root.val) + self.pathSum(root.right, sum-root.val)
        return [[root.val] + node for node in temp]
    '''

437. Path Sum III

https://leetcode.com/problems/path-sum-iii/

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  
    5   -3
   /     
  3   2   11
 /    
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

Solution

https://discuss.leetcode.com/topic/65100/2-python-solutions-with-detailed-explanation

Very slow. 1438ms

# 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 pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: int
        """
        if not root:
            return 0
        return self.findPath(root, sum) + self.pathSum(root.left, sum) + self.pathSum(root.right, sum)
    
    def findPath(self, root, sum):
        res = 0
        if not root:
            return res
        if root.val == sum:
            res += 1
        res += self.findPath(root.left, sum-root.val)
        res += self.findPath(root.right, sum-root.val)
        return res

99ms

# 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 helper(self, root, target, so_far, cache):
        if root == None:
            return 0
        complement = so_far + root.val - target
        result = 0
        if complement in cache:
            result += cache[complement]
        cache.setdefault(so_far+root.val, 0)
        cache[so_far+root.val] += 1
        result += self.helper(root.left, target, so_far+root.val, cache)
        result += self.helper(root.right, target, so_far+root.val, cache)
        cache[so_far+root.val] -= 1
        return result

    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: int
        """
        return self.helper(root, sum, 0, {0:1})