README.md

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.

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.

Solutions (Python)

1. Recursion

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
        if not root:
            return False
        sum -= root.val
        if not root.left and not root.right:
            return sum == 0
        return self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)

2. Iteration

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
        if not root:
            return False
        root.val = sum - root.val
        nodes = [root]
        while nodes:
            curr = nodes.pop()
            if curr.val == 0 and not curr.left and not curr.right:
                return True
            if curr.left:
                curr.left.val = curr.val - curr.left.val
                nodes.append(curr.left)
            if curr.right:
                curr.right.val = curr.val - curr.right.val
                nodes.append(curr.right)
        return False