# 563.Binary Tree Tilt

**563.Binary Tree Tilt**

难度:Easy

> 给定一个二叉树，计算整个树的坡度。

一个树的节点的坡度定义即为，该节点左子树的结点之和和右子树结点之和的差的绝对值。空结点的的坡度是0。

整个树的坡度就是其所有节点的坡度之和。

```
示例:

输入: 
         1
       /   \
      2     3
输出: 1
解释: 
结点的坡度 2 : 0
结点的坡度 3 : 0
结点的坡度 1 : |2-3| = 1
树的坡度 : 0 + 0 + 1 = 1
注意:

任何子树的结点的和不会超过32位整数的范围。
坡度的值不会超过32位整数的范围。
```

使用递归，需要一个辅助函数计算所有节点和，一个全局变量计算所有坡度和。

```
/**
 * 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 {
private:
    int res=0;
    int getSum(TreeNode* root)
    {
        if(!root) return 0;
        int left, right;
        left=getSum(root->left);
        right=getSum(root->right);
        res+=abs(left - right);
        return left+right+root->val ;
    }
public:
    int findTilt(TreeNode* root) {
       getSum(root);
        return res;
        
        
    }
};
```

> 执行用时 :40ms, 在所有 C++ 提交中击败了65.67%的用户\
> 内存消耗 :21.5MB, 在所有 C++ 提交中击败了99.47%的用户


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