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.
For 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.
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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. | |
For 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: | |
# @param root, a tree node | |
# @param sum, an integer | |
# @return a boolean | |
def hasPathSum(self, root, sum): | |
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) |
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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. | |
For 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 binary tree | |
* public class TreeNode { | |
* int val; | |
* TreeNode left; | |
* TreeNode right; | |
* TreeNode(int x) { val = x; } | |
* } | |
*/ | |
public class Solution { | |
public boolean hasPathSum(TreeNode root, int sum) { | |
if (root==null){ | |
return false; | |
} | |
if (root.val==sum && root.left==null && root.right==null){ | |
return true; | |
} | |
return hasPathSum(root.left, sum-root.val)||hasPathSum(root.right, sum-root.val); | |
} | |
} |
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