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: 9 1 6 2 3 -1 -1 -1 4 5 -1 -1 -1 7 -1 -1 8 -1 -1 73 45 11 58 82 25 67 38 42 Sample Output: 58 25 82 11 38 67 45 73 42按照输入建一个数，然后dfs依次赋值，最后bfs输出。
#include<iostream> #include<vector> #include<algorithm> #include<queue> using namespace std; typedef struct node { int num; int value; int l,r; }; node a[100]; int s; int va[100]; queue <node> k; void dfs( int n) { if(a[n].l!=-1) dfs(a[n].l); a[n].num=s; a[n].value=va[s]; s++; if(a[n].r!=-1) dfs(a[n].r); return ; } int main() { s=0; int n; cin>>n; for(int i=0;i<n;i++) { cin>>a[i].l>>a[i].r; } for(int i=0;i<n;i++) cin>>va[i]; sort(va,va+n); dfs(0); k.push(a[0]); node p; bool flag =true; while(!k.empty()) { p=k.front(); k.pop(); if(flag) { flag=false; cout<<p.value; } else cout<<" "<<p.value; if(p.l!=-1) k.push(a[p.l]); if(p.r!=-1) k.push(a[p.r]); } return 0; }
