一、题目
Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However, if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in "zigzagging order" -- that is, starting from the root, print the numbers level-by-level, alternating between left to right and right to left. For example, for the following tree you must output: 1 11 5 8 17 12 20 15.
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 inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print the zigzagging sequence of the tree in a line. 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:
8
12 11 20 17 1 15 8 5
12 20 17 11 15 8 5 1
Sample Output:
1 11 5 8 17 12 20 15
二、题目大意
中序后序建树,正反序层次遍历。
三、考点
TREE
四、注意
1、后序中序建树;
2、queue层序遍历无法保存深度信息,使用dfs保存深度,vector<int> dep[N]。
五、代码
#include<iostream>
#include<vector>
#include<queue>
#define N 32
using namespace std;
struct node {
int val;
struct node *left, *right;
};
vector<int> in, post,level,pre,zig;
vector<int> dep[N];
int max_dep;
node* buildTree(int pl, int pr, int il, int ir) {
if (pl > pr)
return NULL;
node *root = new node();
root->val = post[pr];
root->left = root->right = NULL;
//pre-order
pre.push_back(post[pr]);
//child-tree
int index = il;
while (index <= ir && in[index] != post[pr])
index++;
root->left = buildTree(pl, index-1-il+pl, il, index - 1);
root->right = buildTree(pr-1-(ir-index-1), pr - 1, index + 1, ir);
return root;
}
void dfs(node *root,int depth) {
if (root == NULL)
return;
dep[depth].push_back(root->val);
if (depth > max_dep)
max_dep = depth;
//next
if (root->left != NULL)
dfs(root->left, depth + 1);
if (root->right != NULL)
dfs(root->right, depth + 1);
}
int main() {
//read
int n;
cin >> n;
in.resize(n), post.resize(n);
for (int i = 0; i < n; ++i)
cin >> in[i];
for (int i = 0; i < n; ++i)
cin >> post[i];
//buildTree
node *root = buildTree(0, n - 1, 0, n - 1);
//level-order
queue<node*> que;
que.push(root);
while (!que.empty()) {
node *root = que.front();
level.push_back(root->val);
que.pop();
if (root->left != NULL)
que.push(root->left);
if (root->right != NULL)
que.push(root->right);
}
/*for (int i = 0; i < level.size(); ++i) {
if (i != 0)
cout << " ";
cout << level[i];
}*/
//zig level-order
dfs(root, 1);
cout << dep[1][0];
for (int i = 2; i <= max_dep; ++i) {
if (i % 2 == 0) {
for (int j = 0; j < dep[i].size(); ++j) {
cout <<" "<< dep[i][j];
}
}
else {
for (int j = dep[i].size()-1; j >= 0; --j) {
cout << " " << dep[i][j];
}
}
}
system("pause");
return 0;
}