Path Sum

Question

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.

Tags

• Tree
• DFS

Thought

Use DFS for searching the path. Usually there are two ways to implement DFS:

• stack
• recursion

The implementation for both of them has provided below.

Code

stack for DFS

# 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):
        """
        Use DFS algorithm: stack
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if root is None:
            return False
        stack = [(0, root)]
        while stack:
            acc, ptr = stack.pop()
            if ptr.left is not None:
                stack.append((acc + ptr.val, ptr.left))
            if ptr.right is not None:
                stack.append((acc + ptr.val, ptr.right))
            if ptr.left is None and ptr.right is None:
                if acc + ptr.val == sum:
                    return True
        return False

recursion for DFS

# 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
        if not root.left and not root.right and root.val == sum:
            return True
        sum -= root.val
        return self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)