563.二叉树的坡度（递归）

一、题目

563. 二叉树的坡度 - 力扣（LeetCode）

 

二、代码

 

/**
 * 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 Degree(TreeNode* root, int& Total )
    {       
         if(root == nullptr)
        {
            return 0;
        }
        int leftDegree = Degree(root->left,Total);
        int rightDegree = Degree(root->right,Total);
        int x=abs(leftDegree-rightDegree);//当前结点的坡度

        Total +=x;
        return leftDegree+rightDegree+root->val;
    }
    int findTilt(TreeNode* root) {
        int Total =0;
        Degree(root,Total);
        return Total;
    }
};