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;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param root TreeNode类 
     * @param sum int整型 
     * @return bool布尔型
     */
    bool hasPathSum(TreeNode* root, int sum) {
        if(root == nullptr){
            return false;
        }
        sum -= root->val;
        if(!root->left && !root->right){
            if(sum == 0){
                return true;
            }
            else{
                return false;
            }
        }
        return hasPathSum(root->left, sum) || hasPathSum(root->right, sum);
    }
};