由于太懒 (太菜)不太会链表,就总结一下用数组做二叉树的题

1.已知前序遍历和中序遍历建树

void dfs(int roof,int start,int end,int index){
	if(start>end) return;
	int i=start;
	while(i<end&&in[i]!=pre[roof]) i++;
	lever[index]=pre[roof];
	dfs(roof+1,start,i-1,2*index+2);
	dfs(roof+1+i-start,i+1,end,2*index+1);
}

2.已知中序和后序遍历建树

int build(int l1,int r1,int l2,int r2){
	if(l1>r1) return 0;
	int p1=post[r2];
	int pos=l1;
	while(in[pos]!=post[r2]) pos++;
	int cnt=pos-l1;
	ltree[p1]=build(l1,pos-1,l2,l2+cnt-1);//建左子树 
	rtree[p1]=build(pos+1,r1,l2+cnt,r2-1);//建右子树 
	return p1; 
}

3.层序遍历

void lever_read(int n){
	queue<int>q;
	int p=post[n];
	cout<<p;
	if(ltree[p]) q.push(ltree[p]);
	if(rtree[p]) q.push(rtree[p]);
	while(!q.empty()){
		int pos=q.front();
		q.pop();
		cout<<" "<<pos;
		if(ltree[pos]) q.push(ltree[pos]);
		if(rtree[pos]) q.push(rtree[pos]);
	}
}

7-3 玩转二叉树

已知前序遍历和中序遍历输出反转后的层序遍历,反过来建树就可以了

#include<bits/stdc++.h>
using namespace std;
int in[300];
int pre[300];
int lever[30000];
void dfs(int roof,int start,int end,int index){
	if(start>end) return;
	int i=start;
	while(i<end&&in[i]!=pre[roof]) i++;
	lever[index]=pre[roof];
	dfs(roof+1,start,i-1,2*index+2);
	dfs(roof+1+i-start,i+1,end,2*index+1);
}
int main(){
	int n;
	cin>>n;
	for(int i=0;i<30000;i++){
		lever[i]=-1;
	}
	for(int i=0;i<n;i++){
		cin>>in[i];
	}
	for(int i=0;i<n;i++){
		cin>>pre[i];
	}
	dfs(0,0,n-1,0);
	int cnt=0;
	for(int i=0;i<30000;i++){
		if(cnt!=n-1&&lever[i]!=-1){
			cout<<lever[i]<<" ";
			cnt++;
		}
		else if(lever[i]!=-1){
			cout<<lever[i]<<endl;
		}
	}
}

7-10 树的遍历

已知后序遍历和中序遍历,输出层序遍历

#include<bits/stdc++.h>
using namespace std;
int post[35];
int in[35];
int ltree[35];
int rtree[35];
int build(int l1,int r1,int l2,int r2){
	if(l1>r1) return 0;
	int p1=post[r2];
	int pos=l1;
	while(in[pos]!=post[r2]) pos++;
	int cnt=pos-l1;
	ltree[p1]=build(l1,pos-1,l2,l2+cnt-1);//建左子树 
	rtree[p1]=build(pos+1,r1,l2+cnt,r2-1);//建右子树 
	return p1; 
}
void lever_read(int n){
	queue<int>q;
	int p=post[n];
	cout<<p;
	if(ltree[p]) q.push(ltree[p]);
	if(rtree[p]) q.push(rtree[p]);
	while(!q.empty()){
		int pos=q.front();
		q.pop();
		cout<<" "<<pos;
		if(ltree[pos]) q.push(ltree[pos]);
		if(rtree[pos]) q.push(rtree[pos]);
	}
}
int main(){
	int n;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>post[i];
	}
	for(int i=1;i<=n;i++){
		cin>>in[i];
	}
	build(1,n,1,n);
	lever_read(n);//后序遍历从最后一个节点是根节点 
}