Share my 3 lines c++ solution


  • 11
    Y
    /**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    class Solution {
    public:
        bool hasPathSum(TreeNode* root, int sum) {
            if(!root) return false;
            if((root -> val == sum) && ((!root -> left) && (!root -> right))) return true;
            return hasPathSum(root -> left, sum - root -> val) || hasPathSum(root -> right, sum - root -> val);
        }
    };

  • 1
    D

    it's really a clear answer,I learned that


  • 0
    Y

    thanks!!!!!!!!


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