PAT-A1110. 二叉树-完全二叉树的判断
2017-03-20 21:00
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题目链接:https://www.patest.cn/contests/pat-a-practise/1110
Given a tree, you are supposed to tell if it is a complete binary tree.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=20) which is the total number of nodes in the tree – and hence the nodes are numbered from 0 to N-1. Then N lines follow, each corresponds to a node, and gives the indices of the left and right children of the node. If the child does not exist, a “-” will be put at the position. Any pair of children are separated by a space.
Output Specification:
For each case, print in one line “YES” and the index of the last node if the tree is a complete binary tree, or “NO” and the index of the root if not. There must be exactly one space separating the word and the number.
Sample Input 1:
Sample Output 1:
Sample Input 2:
Sample Output 2:
完全二叉树判断思路:层序遍历,一个完全二叉树中,当遍历到某个结点没有子结点时,那么它后面的所有结点都应该没有子结点。
Given a tree, you are supposed to tell if it is a complete binary tree.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=20) which is the total number of nodes in the tree – and hence the nodes are numbered from 0 to N-1. Then N lines follow, each corresponds to a node, and gives the indices of the left and right children of the node. If the child does not exist, a “-” will be put at the position. Any pair of children are separated by a space.
Output Specification:
For each case, print in one line “YES” and the index of the last node if the tree is a complete binary tree, or “NO” and the index of the root if not. There must be exactly one space separating the word and the number.
Sample Input 1:
9 7 8 - - - - - - 0 1 2 3 4 5 - - - -
Sample Output 1:
YES 8
Sample Input 2:
8 - - 4 5 0 6 - - 2 3 - 7 - - - -
Sample Output 2:
NO 1
完全二叉树判断思路:层序遍历,一个完全二叉树中,当遍历到某个结点没有子结点时,那么它后面的所有结点都应该没有子结点。
#include <iostream> #include <queue> #include <cstring> #include <string> using namespace std; struct BTree { int lchild,rchild; }; BTree T[25]; bool notRoot[25]; int thisone = -1; // 层序遍历时当前访问的节点序数 bool LevelOrder(int root) { queue<int> qt; qt.push(root); bool nochild = false; while (!qt.empty()) { thisone = qt.front(); qt.pop(); if(T[thisone].lchild!=-1){ if(nochild) return false; qt.push(T[thisone].lchild); } else nochild = true; if(T[thisone].rchild!=-1){ if(nochild) return false; qt.push(T[thisone].rchild); } else nochild = true; } return true; } int GetRootIndex() { int i = 0; while (notRoot[i]) { ++i; } return i; } int main() { memset(notRoot, 0, sizeof(notRoot)); int num; cin >> num; for(int i = 0; i < num; ++i) { char str[3]; int temp; scanf("%s", str); if(str[0] == '-') T[i].lchild = -1; else { sscanf(str, "%d", &temp); T[i].lchild = temp; notRoot[temp] = true; } scanf("%s", str); if(str[0] == '-') T[i].rchild = -1; else { sscanf(str, "%d", &temp); T[i].rchild = temp; notRoot[temp] = true; } } int rootIndex = GetRootIndex(); if(LevelOrder(rootIndex)) cout << "YES" << " " << thisone << endl; else cout << "NO" << " " << rootIndex << endl; return 0; }
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