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+ 1064 Complete Binary Search Tree (30分)
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+ < p > 主要是一开始没弄清题目中完全二叉树的定义</ p >
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+ < p > 思路:</ p >
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+ < p > 一棵排序二叉树的中序遍历就是这一组数的递增序列。这边是完全二叉树,假设从0开始,那么节点i的左孩子的标号就是2< em > i+1,右孩子的标号就是2</ em > (i+1)。先将这组数按照递增来排序,然后用中序遍历复原这棵完全排序二叉树,最后直接输出。</ p >
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+ < blockquote >
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+ < p > 完全二叉树:任何一个节点编号x,则左孩子节点为2x,右孩子节点为2x+1< br > 可以使用一个数组存储,从1到n< br > 结论:该数组中元素存放的顺序为该二叉树的层序遍历序列</ p >
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+ < p > 利用完全二叉树建立结构,中序遍历填入元素</ p >
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+ < figure class ="highlight plain "> < table > < tr > < td class ="gutter "> < pre > < span class ="line "> 1</ span > < br > < span class ="line "> 2</ span > < br > < span class ="line "> 3</ span > < br > < span class ="line "> 4</ span > < br > < span class ="line "> 5</ span > < br > < span class ="line "> 6</ span > < br > < span class ="line "> 7</ span > < br > < span class ="line "> 8</ span > < br > < span class ="line "> 9</ span > < br > < span class ="line "> 10</ span > < br > < span class ="line "> 11</ span > < br > < span class ="line "> 12</ span > < br > < span class ="line "> 13</ span > < br > < span class ="line "> 14</ span > < br > < span class ="line "> 15</ span > < br > < span class ="line "> 16</ span > < br > < span class ="line "> 17</ span > < br > < span class ="line "> 18</ span > < br > < span class ="line "> 19</ span > < br > < span class ="line "> 20</ span > < br > < span class ="line "> 21</ span > < br > < span class ="line "> 22</ span > < br > < span class ="line "> 23</ span > < br > < span class ="line "> 24</ span > < br > < span class ="line "> 25</ span > < br > < span class ="line "> 26</ span > < br > < span class ="line "> 27</ span > < br > </ pre > </ td > < td class ="code "> < pre > < span class ="line "> #include <bits/stdc++.h></ span > < br > < span class ="line "> using namespace std;</ span > < br > < span class ="line "> const int maxn = 1000 + 10;</ span > < br > < span class ="line "> </ span > < br > < span class ="line "> int a[maxn], ans[maxn], cnt, N, i;</ span > < br > < span class ="line "> void in_order(int root) {</ span > < br > < span class ="line "> if (root > N)</ span > < br > < span class ="line "> return;</ span > < br > < span class ="line "> in_order(root * 2);</ span > < br > < span class ="line "> ans[root] = a[cnt++];</ span > < br > < span class ="line "> in_order(root * 2 + 1);</ span > < br > < span class ="line "> }</ span > < br > < span class ="line "> </ span > < br > < span class ="line "> int main(int argc, char const *argv[])</ span > < br > < span class ="line "> {</ span > < br > < span class ="line "> cin >> N;</ span > < br > < span class ="line "> for (i = 0; i < N; ++i)</ span > < br > < span class ="line "> cin >> a[i];</ span > < br > < span class ="line "> sort(a, a + N);</ span > < br > < span class ="line "> cnt = 0;</ span > < br > < span class ="line "> in_order(1);</ span > < br > < span class ="line "> for (i = 1; i <= N - 1; ++i)</ span > < br > < span class ="line "> cout << ans[i] << " ";</ span > < br > < span class ="line "> cout << ans[i] << endl;</ span > < br > < span class ="line "> </ span > < br > < span class ="line "> return 0;</ span > < br > < span class ="line "> }</ span > < br > </ pre > </ td > </ tr > </ table > </ figure >
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