# 112. Path Sum
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.
# Solution
Approach 1: BFS/DFS iteration.
# Code (Python)
Approach 1:
class Solution:
def hasPathSum(self, root, sum):
if not root:
return False
stack = [(root, root.val)]
while stack:
node, sum_so_far = stack.pop()
if not node.left and not node.right and sum_so_far == sum:
return True
if node.left:
stack.append((node.left, sum_so_far + node.left.val))
if node.right:
stack.append((node.right, sum_so_far + node.right.val))
return False
# Code (C++)
Approach 1:
/**
* 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;
queue<pair<TreeNode*,int>> nodeQueue;
nodeQueue.push(pair<TreeNode*,int>(root, sum));
while (!nodeQueue.empty())
{
TreeNode *node = nodeQueue.front().first;
int sumRemains = nodeQueue.front().second;
nodeQueue.pop();
if (node->left == NULL && node->right == NULL && node->val == sumRemains)
return true;
if (node->left)
nodeQueue.push(pair<TreeNode*,int>(node->left, sumRemains - node->val));
if (node->right)
nodeQueue.push(pair<TreeNode*,int>(node->right, sumRemains - node->val));
}
return false;
}
};