【PAT】1099. Build A Binary Search Tree (30)
2666 ワード
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:
Sample Output:
分析:
(1)「固定構造」の二叉探索ツリーを与え、ツリー内の配列を与え、このツリーの復元を要求する.
(2)配列を小さいものから大きいものに並べることは,ちょうど二叉探索木の中順遍歴結果であるため,これにより木全体を復元することができる.
(3)木が復元された後,キューを用いて容易に階層を巡ることができる.
この問題の鍵はやはり上述の第2点にある.
正しいコードは次のとおりです.
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分析:
(1)「固定構造」の二叉探索ツリーを与え、ツリー内の配列を与え、このツリーの復元を要求する.
(2)配列を小さいものから大きいものに並べることは,ちょうど二叉探索木の中順遍歴結果であるため,これにより木全体を復元することができる.
(3)木が復元された後,キューを用いて容易に階層を巡ることができる.
この問題の鍵はやはり上述の第2点にある.
正しいコードは次のとおりです.
#include
#include
#include
#include
using namespace std;
vector val;
int index = 0;
struct node{
int val;
int left;
int right;
};
bool cmp(int a, int b){
return a tree;
bool inOrder(node& root){
if(root.left != -1){
inOrder( tree[root.left] );
}
root.val = val[index++];
if(root.right != -1){
inOrder(tree[root.right]);
}
}
bool levelOrder(node root){
queue que;
que.push(root);
bool isFirst = true;
while(!que.empty()){
node t = que.front();
if(isFirst){
cout<