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 path5->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);
            }
        }
    }

};