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.
Example:
Given the below binary tree and sum = 22,
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| 5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
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return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
Solution
(Path-Sum.java) download
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| public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null)
return false;
if (root.left==null && root.right==null)
if (root.val == sum)
return true;
else
return false;
return hasPathSum(root.left,sum-root.val) || hasPathSum(root.right,sum-root.val);
}
}
|
(Path-Sum.py) download
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| # 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.left == None and root.right == None and sum-root.val == 0:
return True
return self.hasPathSum(root.left,sum-root.val) | self.hasPathSum(root.right,sum-root.val)
|