题解 | #二叉树中和为某一值的路径(一)#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param root TreeNode类 
 * @param sum int整型 
 * @return bool布尔型
 */
bool hasPathSum(struct TreeNode* root, int sum ) {
    // write code here
    return totalSum(root,0,sum);
}

int totalSum(struct TreeNode* root,int total,int sum){
    if (root == NULL) {
    return 0;
    }
    if (root->left == NULL&& root->right == NULL ) {
         total += root->val;
         if (total == sum) {
         return 1;
         }
         return 0;
    }
    total += root->val;
    return totalSum(root->left, total,sum) || totalSum(root -> right,total,sum);
}