PAT 甲級 1099 Build A Binary Search Tree
https://pintia.cn/problem-sets/994805342720868352/problems/994805367987355648
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 (≤) which is the total number of nodes in the tree. The next Nlines each contains the left and the right children of a node in the format left_index right_index
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
程式碼:
#include <bits/stdc++.h> using namespace std; int N; struct Node{ int data; int l, r; }node[110]; vector<int> v; int ans = 0; void inorder(int root) { if(root == -1) return ; inorder(node[root].l); node[root].data = v[ans ++]; inorder(node[root].r); } void levelorder(int root) { queue<int> q; if(root != -1) q.push(root); bool isfirst = true; while(!q.empty()) { int t = q.front(); q.pop(); if(isfirst) { printf("%d", node[t].data); isfirst = false; } else printf(" %d", node[t].data); if(node[t].l != -1) q.push(node[t].l); if(node[t].r != -1) q.push(node[t].r); } } int main() { scanf("%d", &N); for(int i =0; i < N; i ++) scanf("%d%d", &node[i].l, &node[i].r); v.resize(N); for(int i = 0; i < N; i ++) scanf("%d", &v[i]); sort(v.begin(), v.end()); inorder(0); levelorder(0); return 0; }
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