跟以前那类题目做法相同，对于这棵树，进行dfs,同时记录当前层数距离跟的关系，用奇偶数来表示，然后再以各个节点被dfs的时间戳 来建立区间 让树状数组映射上去，最后奇偶分开，加的在一个树状数组里，减去的在另一个里面，然后 最后求单点值的时候 就是自己这个点 所属的 距离根节点的关系，也就是自己应该加上的值，再减去对应的另一个树状数组里的应该减去的值，然后 一开始 各个节点本身具有的值 并没有加进树状数组里，还得加上原本具有的值，这样就是答案了

然后又用线段树做了一下，也是以时间戳来搞，同时记录这个节点距离根的奇偶性，然后也是建立两颗线段树，一个记录奇数处理，一个记录偶数处理，结果不知哪里写错了，又改了很久，不行又重新写了一下，真是学啥忘啥。。

树状数组的：

int n;int m;int c[2][200000 * 2 + 55];typedef struct Node {	int l,r,val;	int now;};Node node[200000 + 55];vector G[200000 + 55];int cnt;void init() {	memset(c,0,sizeof(c));	for(int i=0;i<200000 + 55;i++)G[i].clear();}bool input() {	while(cin>>n>>m) {	for(int i=1;i<=n;i++)cin>>node[i].val;	int tmp = n - 1;	while(tmp--) {	int u,v;	scanf("%d %d",&u,&v);	G[u].push_back(v);	G[v].push_back(u);	}	return false;	}	return true;}int lowbit(int x) {	return x&(-x);}void add(int i,int val,int *aa) {	while(i <= 2 * n) {	aa[i] += val;	i += lowbit(i);	}}int get_sum(int i,int *aa) {	int sum = 0;	while(i > 0) {	sum += aa[i];	i -= lowbit(i);	}	return sum;}void dfs(int u,int pre,int tot) {	node[u].l = cnt++;	node[u].now = tot;	for(int i=0;i>type;	if(type == 1) {	int x,y;	cin>>x>>y;	//int tmp = node[x].now;	//int aa = node[x].l;	//int bb = node[x].r;	add(node[x].l,y,c[node[x].now]);	add(node[x].r + 1,-y,c[node[x].now]);	}	else {	int x;	cin>>x;	//int aa = (get_sum(node[x].l,c[node[x].d]) /*- get_sum(node[x].l - 1,c[node[x].d])*/);	//int bb = (get_sum(node[x].l,c[node[x].d^1])/* - get_sum(node[x].l - 1,c[node[x].d^1])*/);	//int cc = 0;	int ans = get_sum(node[x].l,c[node[x].now]) - get_sum(node[x].l,c[node[x].now^1]);	ans += node[x].val;	cout< 
 
 
线段树的：
 
const int N = 200000 + 55;int n;int m;int nnum[N + 55];int le[N + 55],ri[N + 55],belong[N + 55];int head[N + 55];typedef struct Node {	int l,r;	ll sum;	int lazy;};Node tree_even[N * 4 + 55],tree_odd[N * 4 + 55];typedef struct NODE {	int fro,to;	int nex;};NODE edge[2 * N + 55];int tot;int cnt;void add(int u,int v) {	edge[tot].fro = u;	edge[tot].to = v;	edge[tot].nex = head[u];	head[u] = tot++;}void dfs(int u,int pre,int d) {	le[u] = ++cnt;	for(int i=head[u];i!=-1;i=edge[i].nex) {	int v = edge[i].to;	if(v == pre)continue;	dfs(v,u,d^1);	}	belong[le[u]] = d;	ri[le[u]] = cnt;}void push_up(int id,Node *tree) {	tree[id].sum = tree[id<<1].sum + tree[id<<1|1].sum;}void push_down(int id,Node *tree) {	if(tree[id].lazy == 0)return ;	tree[id].sum += (tree[id].r - tree[id].l + 1) * tree[id].lazy;	if(tree[id].l == tree[id].r) {	tree[id].lazy = 0;	return ;	}	tree[id<<1].lazy += tree[id].lazy;	tree[id<<1|1].lazy += tree[id].lazy;	tree[id].lazy = 0;}void build(int l,int r,int id,Node *tree) {	tree[id].l = l;	tree[id].r = r;	tree[id].lazy = 0;	if(l == r) {	tree[id].sum = 0ll;	return ;	}	int mid = (l + r)>>1;	build(l,mid,id<<1,tree);	build(mid + 1,r,id<<1|1,tree);	push_up(id,tree);}void update(int l,int r,int id,ll val,Node *tree) {	if(tree[id].l == l && tree[id].r == r) {	tree[id].lazy += val;	push_down(id,tree);	return ;	}	push_down(id,tree);	int mid = (tree[id].l + tree[id].r)>>1;	if(r <= mid)update(l,r,id<<1,val,tree);	else if(l > mid)update(l,r,id<<1|1,val,tree);	else {	update(l,mid,id<<1,val,tree);	update(mid + 1,r,id<<1|1,val,tree);	}	push_up(id,tree);}ll query(int l,int r,int id,Node *tree) {	if(tree[id].l == l && tree[id].r == r) {	push_down(id,tree);	return tree[id].sum;	}	push_down(id,tree);	int mid = (tree[id].l + tree[id].r)>>1;	ll ret = 0ll;	if(r <= mid)ret += query(l,r,id<<1,tree);	else if(l > mid)ret += query(l,r,id<<1|1,tree);	else {	ret += query(l,mid,id<<1,tree);	ret += query(mid + 1,r,id<<1|1,tree);	}	return ret;}void init() {	memset(tree_even,0,sizeof(tree_even));	memset(tree_odd,0,sizeof(tree_odd));	memset(head,-1,sizeof(head));	tot = 1;	cnt = 0;}bool input() {	while(cin>>n>>m) {	for(int i=1;i<=n;i++)cin>>nnum[i];	for(int i=1;i>u>>v;	add(u,v);	add(v,u);	}	return false;	}	return true;}void cal() {	dfs(1,-1,1);	build(1,n,1,tree_even);	build(1,n,1,tree_odd);	while(m--) {	int type;	cin>>type;	if(type == 1) {	int x,y;	cin>>x>>y;	int left = le[x];	int right = ri[left];	if(belong[left]&1) update(left,right,1,y,tree_odd);	else update(left,right,1,y,tree_even);	}	else {	int x;	cin>>x;	int left = le[x];	ll ans;	if(belong[left]&1)	ans = query(left,left,1,tree_odd) - query(left,left,1,tree_even);	else	ans = query(left,left,1,tree_even) - query(left,left,1,tree_odd);	ans += nnum[x];	cout< 




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