7-1 玩转二叉树

Posted 2021-02-09 bxd123

给定一棵二叉树的中序遍历和前序遍历，请你先将树做个镜面反转，再输出反转后的层序遍历的序列。所谓镜面反转，是指将所有非叶结点的左右孩子对换。这里假设键值都是互不相等的正整数。

 

先是建树 快速套用模板

然后  他是按照层遍历  那就用bfs  如果按照某种序遍历  用dfs

#include<bits/stdc++.h>
using namespace std;
    int n;
    int zhong[35];
    int qian[35];
    int le[35],ri[35];
int built(int x1,int y1,int x2,int y2)
{
    if(x2>y2||x1>y1)return 0;
    int root=qian[x1];
    int p=x2;
    while(zhong[p]!=root)p++;
    int cnt=p-x2;
     ri[root]=built(x1+cnt+1,y1,x2+cnt+1,y2);
     le[root]=built(x1+1,x1+cnt,x2,x2+cnt-1);
    return root;
}
void bfs(int root)
{
    queue<int >q;
    q.push(root);
    printf("%d",root);
    int first=1;
    while(!q.empty())
    {
        int u=q.front();q.pop();
        if(first)first=0;
        else
        printf(" %d",u);
        if(ri[u]!=0)q.push(ri[u]);
        if(le[u]!=0)q.push(le[u]);
    }
}

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&zhong[i]);
    for(int i=1;i<=n;i++)scanf("%d",&qian[i]);
    int root=qian[1];
    built(1,n,1,n);
    bfs(root);
    return 0;
}