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.
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->2which sum is 22.The Idea:
- Iterative through the tree and back track when you need to. Keep a global variable that will help you exit from all the recursive call stack.
class Solution {
public:
bool isFound = false;
bool hasPathSum(TreeNode* root, int sum) {
hasPathSum_rec(root, sum);
return isFound;
}
void hasPathSum_rec(TreeNode *root, int sum) {
if (root && !isFound) {
sum = sum - root->val;
if (!root->left && !root->right && sum == 0)
isFound = true;
else {
hasPathSum(root->left, sum);
hasPathSum(root->right, sum);
}
}
}
};