[LeetCode]563. Binary Tree Tilt(二叉树的差值)

563. Binary Tree Tilt

Given a binary tree, return the tilt of the whole tree.

The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0.

The tilt of the whole tree is defined as the sum of all nodes’ tilt.
给出一个二叉树，返回整个树的差值。
树节点的差值被定义为所有左子树节点值和所有右子树节点值之和的差的而绝对值。 空节点有差值0。
整个树的差值被定义为所有节点的差值的总和。
Example:

Input: 
     1
   /   \
  2     3
Output: 1
Explanation: 
Tilt of node 2 : 0
Tilt of node 3 : 0
Tilt of node 1 : |2-3| = 1
Tilt of binary tree : 0 + 0 + 1 = 1

Note:

1. The sum of node values in any subtree won’t exceed the range of 32-bit integer.
2. All the tilt values won’t exceed the range of 32-bit integer.

思路：
这道题其实是要求 求出各个节点左右子树的差的绝对值，将这些绝对值求和并返回。
左右子树的差 = | 左子树所有节点的值的和 - 右子树所有节点的值的和 |
(刚开始没好好看题，只算了各个节点左右孩子的差，导致一直出错)

代码如下：

class Solution {
public:
    // Definition for a binary tree node.
    struct TreeNode {
        int val;
        TreeNode *left;
        TreeNode *right;
        TreeNode(int x) : val(x), left(NULL), right(NULL) {}
    };
    int tilt = 0;
    int findTilt(TreeNode* root) {
        sum(root);
        return tilt;
    }
    int sum(TreeNode* root){
        if(root == nullptr)
            return 0;
        int left = sum(root->left);//左子树各节点和
        int right = sum(root->right);//右子树各节点和
        tilt += left>right ? left-right : right-left;
        //tilt += abs(left - right);
        return left + right + root->val;//返回左右子树各节点和
    }
};

注意：

1. 求绝对值也可以使用函数abs()，要记得加头文件#include