1099. Build A Binary Search Tree (30)建立二叉搜索树

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

The left subtree of a node contains only nodes with keys less than the node’s key.

The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.

Both the left and right subtrees must also be binary search trees.

Given the structure of a binary tree and a sequence of distinct integer keys, there is only one way to fill these keys into the tree so that the resulting tree satisfies the definition of a BST. You are supposed to output the level order traversal sequence of that tree. The sample is illustrated by Figure 1 and 2.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=100) which is the total number of nodes in the tree. The next N lines each contains the left and the right children of a node in the format “left_index right_index”, provided that the nodes are numbered from 0 to N-1, and 0 is always the root. If one child is missing, then -1 will represent the NULL child pointer. Finally N distinct integer keys are given in the last line.

Output Specification:

For each test case, print in one line the level order traversal sequence of that tree. All the numbers must be separated by a space, with no extra space at the end of the line.

Sample Input:

91 62 3-1 -1-1 45 -1-1 -17 -1-1 8-1 -173 45 11 58 82 25 67 38 42

Sample Output:

58 25 82 11 38 67 45 73 42

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#include<iostream>#include<algorithm>#include<vector>using namespace std;struct tree{int num,left,right;};struct tree T[10000];int num[10000];int index1=0;int max_lev=1;vector<int > leval[1000];void fun1(int root)//按顺序树里填值{    if (T[root].left==-1&&T[root].right==-1)    {        T[root].num=num[index1++];        return ;    }    if(T[root].left!=-1)        fun1(T[root].left);    T[root].num=num[index1++];    if(T[root].right!=-1)        fun1(T[root].right);}void fun2(int root,int lev)//查找每一层的数{    if (max_lev<lev)        max_lev=lev;    leval[lev].push_back(T[root].num);    if(T[root].left!=-1)        fun2(T[root].left,lev+1);    if(T[root].right!=-1)        fun2(T[root].right,lev+1);}int main(){    int n;    cin>>n;    for(int i=0;i<n;i++)        cin>>T[i].left>>T[i].right;    for (int i=0;i<n;i++)        cin>>num[i];    sort(num,num+n);//二叉搜索树的中序遍历是非下降序列。    fun1(0);    fun2(0,1);    for (int i=1;i<=max_lev;i++)    {        for (int j=0;j<leval[i].size();j++)        {            cout<<leval[i][j];            if (j!=leval[i].size()&&i!=max_lev)                cout<<" ";        }    }    return 0;}