LeetCode 814. 二叉树剪枝(DFS统计和)

二叉树剪枝
用check()函数统计某一棵子树的节点值的和，然后，如果左子树为0，将root->left置为0，右子树也是如此；

/**
 * 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:
    TreeNode* pruneTree(TreeNode* root) {
        int sum = check(root);
        if(!sum){
            root = nullptr;
        }
        return root;
    }
    bool check(TreeNode* root){
        if(!root){
            return 0;
        }
        int leftSum = check(root->left) ;
        int rightSum = check(root->right);
        if(!leftSum){
            root->left = nullptr;
        }
        if(!rightSum){
            root->right = nullptr;
        }
        return root->val + leftSum + rightSum;
    }
};