PAT Advanced 1110 Complete Binary Tree

1110 Complete Binary Tree

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

解題思路

給定若干的二叉樹，問是夠能夠組成一顆完全二叉樹， 是的話輸出完全二叉樹中最後一個元素， 如果不是輸出根節點。 符合是完全二叉樹的條件是從根節點遍歷，到最後元素的下標值等於節點個數值。

輸入方式： 行數代表根節點，行數輸出的值爲根節點元素，如果如果爲空，用-帶替。

過程：

1. 找到根節點，二叉樹的性質中根節點不會作爲子節點， 圖示中的根節點就是6，通過hash的方法找到根節點。
2. 遍歷選擇dfs，這樣方便找對應節點的下標。
3. 判斷是否是完全二叉樹， 如果是完全二叉樹的話節點個數和最後一個節點的下標數相同，如果不是的話，那麼最後一個節點的下標數 > 節點數， 因爲其中存在不連續的下標值，所以最後一個節點的下標數一定比節點個數的值大。

圖示

給定的起始的二叉樹如下：

完成構建二叉樹：

解題代碼

#include <iostream>
using namespace std;
int btree[25][2], a[25];
int n, idx, last;
void dfs(int v, int  cur){
    if (v == -1) return;
    if (idx < cur) idx = cur, last = v;
    dfs(btree[v][0], 2 * cur);
    dfs(btree[v][1], 2 * cur + 1);
}
int main(){
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        string sl, sr;
        cin >> sl >> sr;
        btree[i][0] = btree[i][1] = -1;
        if (sl != "-") btree[i][0] = stoi(sl), a[stoi(sl)] = 1;
        if (sr != "-") btree[i][1] = stoi(sr), a[stoi(sr)] = 1;
    }
    int root;
    for (int i = 0; i < n; i++) if (!a[i]) root = i;
    dfs(root, 1);
    if (idx == n) printf("YES %d", last);
    else printf("NO %d", root);
    return 0;
}