1099. Build A Binary Search Tree (30)
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
思路
1.1064的老办法,将节点值升序排序后就是搜索树的中序遍历序列。
2.从根节点开始按中序遍历构造整棵树。
3.层次遍历打印整棵树(用队列BFS)。
代码
#include<iostream> #include<vector> #include<algorithm> #include<queue> using namespace std; class Node { public: int left; int right; int val; }; vector<Node> bstnodes(101); vector<int> nodevalue(101); int index; void createBST(int root) { if(bstnodes[root].left != -1) createBST(bstnodes[root].left); bstnodes[root].val = nodevalue[index++]; if(bstnodes[root].right != -1) createBST(bstnodes[root].right); } int main() { int N; while(cin >> N) { index = 0; for(int i = 0;i < N;i++) { cin >> bstnodes[i].left >> bstnodes[i].right; } for(int i = 0;i < N;i++) { cin >> nodevalue[i]; } sort(nodevalue.begin(),nodevalue.begin()+N); createBST(0); //print queue<int> q; q.push(0); while(!q.empty()) { int temp = q.front(); q.pop(); if(temp != 0) cout <<" "; cout << bstnodes[temp].val; if(bstnodes[temp].left != -1) q.push(bstnodes[temp].left); if(bstnodes[temp].right != -1) q.push(bstnodes[temp].right); } cout << endl; } return 0; }