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.
给定一二叉树，返回其整棵树的倾斜度。
倾斜度定义为左子树结点和与右子树结点和的绝对值之差。空节点的倾斜度为0。
整棵树的倾斜度定义为所有结点倾斜度的和。

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
后序遍历，遍历过程中递归计算左右子树的结点和，同时使用全局变量记录当前节点倾斜度。

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

或不使用全局变量，改为参数传递。

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