Depth-First Search
Path Sum
easy
Count: 10
DESCRIPTION (inspired by Leetcode.com)
Given the root of a binary tree and an integer target, write a recursive function to determine if the tree has a root-to-leaf path where all the values along that path sum to the target.
Example 1:
Input:
[4, 2, 7, 1, 3, 6, 9] target = 17
Output: true (the path is 4 -> 7 -> 6)
Example 2:
Input:
[4, 2, 7, 1, 3, 6, 9] target = 13
Output: false
Explanation
We can solve this problem using the framework established in the Overview section.
Return Values
If I'm at a node in the tree, what values do I need from my child to tell if there's a path from the root to a leaf that sums to the target?
In order for there to be a path from the root to a leaf that sums to the target, there must be a path from either the left or right child to a leaf that sums to target - node.val.
This tells us that each recursive call should return True if there's a path from the current node to a leaf that sums to the current target, and False otherwise. In the body, we'll make recursive calls to the current node's left and right children, passing in target - node.val for the current target. If either of those calls returns True, then we should return True to the parent.
Base Case
If our current node is None, then our subtree is empty and there's no path from the current node to a leaf that sums to the target.
If our current node is a leaf node, then we check if the target is equal to the leaf node's value. If it is, then there's a path from the current node to a leaf that sums to the target.
Helper Functions
We need to pass the current target down from parent to children nodes, but this is included in the function signature of the main function, so we don't need to introduce a helper function.
Global Variables
The return value of each recursive call matches the answer to the problem, so we don't need to use any global variables.
Solution
Solution
class Solution:def pathSum(self, root, target):if root is None:return False# if we reach a leaf node, check if the target is equal to the leaf node's valueif not root.left and not root.right:return target == root.valtarget -= root.val# check if there's a path from the current node to a leaf that sums to targetreturn self.pathSum(root.left, target) or self.pathSum(root.right, target)
Animated Solution
Visualization
Try these examples:
def hasPathSum(root, target):if not root:return Falseif not root.left and not root.right:return target == root.valleft = hasPathSum(root.left, target - root.val)right = hasPathSum(root.right, target - root.val)return left or right
target: 13
path sum
0 / 13
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Complexity Analysis
Time Complexity: O(N) where N is the number of nodes in the tree. We visit each node exactly once, and each node, we perform a constant amount of work.
Space Complexity: O(N) we have to allocate space for a total of N (one for each nodes) recursive calls on the call stack.
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