力扣-563题 二叉树的坡度（C++）- dfs

题目链接：https://leetcode-cn.com/problems/binary-tree-tilt/
题目如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
    
    
public:
    int result=0;
    int findTilt(TreeNode* root) {
    
    
        //在遍历每个节点时，累加其左子树节点之和与右子树节点之和的差的绝对值，并返回以其为根节点的树的节点之和
        if(root==NULL) return 0;
        dfs(root);

        return result;
    }

    int dfs(TreeNode* root){
    
    //左右根，后序
        if(root==NULL) return 0;

        int left=dfs(root->left);
        int right=dfs(root->right);

        result+=abs(left-right);

        return left+right+root->val;
    }
};