1020 Tree Traversals——PAT甲级真题
1020 Tree Traversals
Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input:
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7Sample Output:
4 1 6 3 5 7 2
题目大意:给你一颗二叉树的后序遍历序列和中序遍历序列。然后求出这颗二叉树的层次遍历序列。
大致思路:首先我们因该知道后序遍历序列的最后一个点为二叉树的根节点,在中序遍历序列中根节点左侧都为左子树,右侧都为右子树。所以我们要首先找到二叉树的根节点,然后在中序遍历中找到根节点所在的位置,然后对二叉树的左右子树递归的调用上述过程。
代码:
#include <bits/stdc++.h>
using namespace std;
const int N = 35;
struct TreeNode {
int val;
TreeNode* left;
TreeNode* right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
int postorder[N], inorder[N]; //后序遍历数组,中序遍历数组
int preorder[N];
int n;
//由后续和中序遍历创建二叉树
TreeNode* buildTree(int postL, int postR, int inL, int inR) {
if (postL > postR)
return NULL; //设置递归返回条件,当postL ==
//postR时表明指向叶子节点,当postl > postR时,应当返回
TreeNode* node = new TreeNode(postorder[postR]);
//查找根节点在中序遍历中的位置
int k;
for (k = inL; k <= inR; k++) {
if (inorder[k] == postorder[postR]) break;
}
int numL = k - inL; //计算左子树结点的个数
//中序遍历中根节点左边的是左子树,右边的是右子树递归建树
node->left = buildTree(postL, postL + numL - 1, inL, k - 1);
node->right = buildTree(postL + numL, postR - 1, k + 1, inR);
return node;
}
//由前序和中序遍历创建二叉树
TreeNode* buildTree2(int preL, int preR, int inL, int inR) {
if (preL > preR) return nullptr;
TreeNode *node = new TreeNode(preorder[preL]);
int k;
for (k = inL; k <= inR; k++) {
if (inorder[k] == preorder[preL]) break;
}
int numL = k - inL;
node->left = buildTree2(preL + 1, preL + numL, inL, k - 1);
node->right = buildTree2(preL + numL + 1, preR, k + 1, inR);
return node;
}
//先序遍历
void inordervisit(TreeNode* root) {
if (root == nullptr) return;
cout << root->val << " ";
inordervisit(root->left);
inordervisit(root->right);
}
//层次遍历
void BFS(TreeNode* root) {
queue<TreeNode*> q;
q.push(root);
int cnt++;
while(!q.empty()) {
auto node = q.front(); q.pop();
cout << node->val;
cnt++;
if (cnt != n) cout << " ";
else cout << endl;
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
}
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &postorder[i]);
for (int i = 1; i <= n; i++) scanf("%d", &inorder[i]);
TreeNode* root = buildTree(1, n, 1, n);
BFS(root);
// inordervisit(root);
return 0;
}