题解 | #二叉树中和为某一值的路径(一)#

DFS求解

 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
private:
    int res = false;
    int cur_sum = 0;
public:
    /**
     * 
     * @param root TreeNode类 
     * @param sum int整型 
     * @return bool布尔型
     */
    bool hasPathSum(TreeNode* root, int sum) {
        // write code here
        dfs(root, sum);
        return res;
    }
    
    void dfs(TreeNode* root, int sum){
        if(root == nullptr){
            return;
        }
        if(res){
            return;
        }
        cur_sum += root->val;
        
        if(cur_sum == sum && root->left == nullptr && root->right == nullptr){
            res = true;
            return;
        }

        dfs(root->left, sum);
        dfs(root->right, sum);
        cur_sum -= root->val;
        
    }
};