题解 | #二叉树中和为某一值的路径(一)#

bool hasPathSum(struct TreeNode* root, int sum ) {
    if(!root) return false;
    if(!root->left&&!root->right&&sum==root->val) return true;
    return hasPathSum(root->left,sum-root->val)||hasPathSum(root->right,sum-root->val);
}