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.
# 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:
def dfs(root, sum):
if root is None:
return False
if root.val == sum and root.left is None and root.right is None:
return True
return dfs(root.left, sum - root.val) or dfs(root.right, sum - root.val)
return dfs(root, sum)/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
return dfs(root, sum);
}
private boolean dfs(TreeNode root, int sum) {
if (root == null) return false;
if (root.val == sum && root.left == null && root.right == null) return true;
return dfs(root.left, sum - root.val) || dfs(root.right, sum - root.val);
}
}