Published: Oct 4, 2022
The pre-order traversal works to find the answer. When the depth-first search (recursion) approach is taken, we should be careful about the value consistency when the process comes back from deeper stack. Other than that, the problem is just a tree traversal.
rootof a binary tree and an integer
targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals
A leaf is a node with no children.
- The number of nodes in the tree is in the range
-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000
Example 1 Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22 Output: true Explanation: 5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1 Path: 5 -> 4 -> 11 -> 2
Example 2 Input: root = [1,2,3], targetSum = 5 Output: false Explanation: 1 / \ 2 3
Example 3 Input: root = , targetSum = 0 Output: false Explanation: The tree is empty.
The solution here takes the pre-order traversal approach (depth-first search). The first step is to subtract the root value from the given targetSum. If the current node is a leaf, check the targetSum becomes zero. The left or right subtree returns true if the targetSum exists.
# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class PathSum: def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool: if not root: return False targetSum -= root.val if not root.left and not root.right: return targetSum == 0 left = self.hasPathSum(root.left, targetSum) right = self.hasPathSum(root.right, targetSum) return left or right