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.
Approach #1: C++.
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool hasPathSum(TreeNode* root, int sum) {
if (root == NULL) return false;
if (root->val == sum && root->right == NULL && root->left == NULL) return true;
return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);
}
};
Approach #2: Java.
/**
* 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) {
if (root == null) return false;
if (root.val == sum && root.right == null && root.left == null) return true;
return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum-root.val);
}
}
Approach #3: Python.
# 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 root == None:
return False
if root.val == sum and root.left == None and root.right == None:
return True
return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)