Leetcode: Path Sum

Path Sum

Similar Problems:

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,
             / \
            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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## Blog link: https://code.dennyzhang.com/path-sum
# 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):
        if root is None:
            return False
        queue = []
        queue.append((root, 0))
        while len(queue) != 0:
            (element, current_sum) = queue[0]
            del queue[0]
            if element.left is None and element.right is None:
                if element.val + current_sum == sum:
                    return True
            if element.left:
                queue.append((element.left, current_sum + element.val))
            if element.right:
                queue.append((element.right, current_sum + element.val))
        return False

    def hasPathSum_v1(self, root, sum):
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        ## Idea: DFS recursive, BFS
        ## Complexity: Time O(n), Space O(log(k))
        return self._hasPathSum(root, sum, 0)

    def _hasPathSum(self, root, sum, current_sum):
        if root is None:
            return False
        if root.left is None and root.right is None:
            return root.val + current_sum == sum

        if root.left:
            if self._hasPathSum(root.left, sum, current_sum + root.val):
                return True

        if root.right:
            if self._hasPathSum(root.right, sum, current_sum + root.val):
                return True

        return False

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