/*** 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 {
public://每一个节点都有接受上传下发的需求，只统计单向的即可//对于单个节点来说，递归int distributeCoins(TreeNode* root) {int ret = 0;if (root == NULL){return 0;}else{int tmpleftnodes = 0;int tmpleftconins = 0;int tmprightnodes = 0;int tmprightcoins = 0;int tmpselfnode = 0;int tmpselfcoins = 0;GetInfo(root, tmpleftnodes, tmpleftconins, tmprightnodes, tmprightcoins, tmpselfnode, tmpselfcoins);int tmpdeltaleft = tmpleftconins - tmpleftnodes;int tmpdeltaright = tmprightcoins - tmprightnodes;//int tmpselfdelta = tmpselfcoins - tmpselfnode;int tmpselfdelta = 0;ret = ret + abs(tmpselfdelta - tmpdeltaleft) + abs(tmpselfdelta - tmpdeltaright);ret = ret + distributeCoins(root->left) + distributeCoins(root->right);}return ret;}//获得该节点左右节点数以及所拥有的硬币数void GetInfo(TreeNode* root, int& leftnodes, int& leftconins, int& rightnodes, int& rightcoins,int& selfnode,int& selfcoins){if (root == NULL){leftnodes = 0;leftconins = 0;rightnodes = 0;rightcoins = 0;selfnode = 0;selfcoins = 0;return;}else if (root->left == NULL && root->right == NULL){leftnodes = 0;leftconins = 0;rightnodes = 0;rightcoins = 0;selfnode = 1;selfcoins = root->val;return;            }else //if (root->left != NULL && root->right != NULL)
        {int tmpleftnodes = 0;int tmpleftconins = 0;int tmprightnodes = 0;int tmprightcoins = 0;int tmpselfnode = 0;int tmpselfcoins = 0;GetInfo(root->left, tmpleftnodes, tmpleftconins, tmprightnodes, tmprightcoins, tmpselfnode, tmpselfcoins);leftnodes = tmpleftnodes + tmprightnodes + tmpselfnode;leftconins = tmpleftconins + tmprightcoins + tmpselfcoins;tmpleftnodes = 0;tmpleftconins = 0;tmprightnodes = 0;tmprightcoins = 0;tmpselfnode = 0;tmpselfcoins = 0;GetInfo(root->right, tmpleftnodes, tmpleftconins, tmprightnodes, tmprightcoins, tmpselfnode, tmpselfcoins);rightnodes = tmpleftnodes + tmprightnodes + tmpselfnode;rightcoins = tmpleftconins + tmprightcoins + tmpselfcoins;selfnode = 1;selfcoins = root->val;}}};

typedef signed long long ll;#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------class Solution {
public:int ret;pair<int,int> count(TreeNode* root) {pair<int,int> A={1,root->val};if(root->left) {auto a=count(root->left);A.first+=a.first;A.second+=a.second;}if(root->right) {auto a=count(root->right);A.first+=a.first;A.second+=a.second;}ret+=abs(A.first-A.second);return A;}int distributeCoins(TreeNode* root) {ret=0;count(root);return ret;}
};

/*** 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 {
public:int sgn(int x) {if (x == 0)return 0;return x < 0 ? -1 : 1;}int maxTurbulenceSize(vector<int>& A) {int n = A.size();int l = 0;int ans = 1;for (int r = 1; r < n; ++r) {if (l + 1 < r and sgn(a[r] - a[r - 1]) == sgn(a[r - 1] - a[r - 2])) {l = r - 1;}if (a[r] == a[r - 1]) {l = r;}ans = max(ans, r - l + 1);}return ans;}
};*/class Solution {
public:int sz(TreeNode* root) {if (root == nullptr)return 0;return 1 + sz(root->left) + sz(root->right);}int coins(TreeNode* root) {if (root == nullptr) {return 0;}return root->val + coins(root->left) + coins(root->right);}int distributeCoins(TreeNode* root) {// number of coins over edge = diff between coins and nodes for each subtreeif (root == nullptr) {return 0;}return abs(coins(root) - sz(root)) + distributeCoins(root->left) + distributeCoins(root->right);}
};