判断是否存在二叉树中和为某一值的路径

Posted 2022-11-25 _BitterSweet

/**
 * 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;
        else if(root -> val == sum && root -> left == nullptr && root -> right == nullptr)
        
            return true;
        
        return hasPathSum(root -> left, sum - root -> val) || hasPathSum(root -> right, sum - root -> val);
    
    
;