二叉树垂直遍历 java_【004】二叉树垂直遍历
二叉树垂直遍历
题目描述
输入输出
示例输入
实例输出
DFS
BFS
更简单的方法
二叉树垂直遍历
题目描述
对于一个二叉树，输出它的垂直遍历结果；对于同一列的节点，按照从左向右，从上向下的顺序排列。
例如，对于以下二叉树：
1
/ \
2 3
/
4
垂直遍历的结果是：2 1 4 3
输入输出
输入
- 第一行是n，表示节点个数(节点编号从0到n-1)；当n=-1时，表示输入结束
- 之后的n行，每一行有三个整数，分别表示：节点的数值，左子树的编号，右子树的编号(编号-1表示节点为空)
输出
- 针对每组输入，输出垂直遍历的结果
示例输入
4
1 1 2
2 -1 -1
3 3 -1
4 -1 -1
-1
实例输出
2 1 4 3
DFS
#include
#include
#include
#include
#include
#include
using namespace std;
struct Node {
int data;
Node *left;
Node *right;
};
class TreePrint {
private:
map > nmap;
public:
void dfs(Node *root, int pos) {
if (root == NULL) return;
nmap[pos].push_back(root->data);
dfs(root->left, pos-1);
dfs(root->right, pos+1);
}
void display() {
int min=0;
while (nmap.find(min) != nmap.end()) --min;
for (int i=min+1; nmap.find(i) != nmap.end(); ++i)
for (vector::iterator it = nmap[i].begin();
it != nmap[i].end(); ++it)
printf("%d ", *it);
printf("\n");
}
};
int main() {
ifstream in("../input.txt");
int n;
in >> n;
while (n != -1) {
Node *tree = new Node[n];
for (int i=0; iint tmp, left, right;
in >> tmp >> left >> right;
tree[i].data = tmp;
if (left != -1) tree[i].left = &tree[left];
else tree[i].left = NULL;
if (right != -1) tree[i].right = &tree[right];
else tree[i].right = NULL;
}
TreePrint tp;
tp.dfs(&tree[0], 0);
tp.display();
in >> n;
}
in.close();
return 0;
}
然而这个结果并不正确，不能妥善的处理孩子节点超过父节点的深度的情况。
BFS
#include
#include
#include
#include
#include
#include
using namespace std;
struct Node {
int data;
Node *left;
Node *right;
};
class TreePrint {
private:
map > nmap;
public:
void dfs(Node *root, int pos) {
if (root == NULL) return;
nmap[pos].push_back(root->data);
dfs(root->left, pos-1);
dfs(root->right, pos+1);
}
void bfs(Node *root) {
queue q;
queue qpos;
q.push(root);
qpos.push(0);
while (!q.empty()) {
Node *tmp = q.front();
int pos = qpos.front();
q.pop();
qpos.pop();
nmap[pos].push_back(tmp->data);
if (tmp->left != NULL) {
q.push(tmp->left);
qpos.push(pos-1);
}
if (tmp->right != NULL) {
q.push(tmp->right);
qpos.push(pos+1);
}
}
}
void display() {
int min=0;
while (nmap.find(min) != nmap.end()) --min;
for (int i=min+1; nmap.find(i) != nmap.end(); ++i)
for (vector::iterator it = nmap[i].begin();
it != nmap[i].end(); ++it)
printf("%d ", *it);
printf("\n");
}
};
int main() {
ifstream in("../input.txt");
int n;
in >> n;
while (n != -1) {
Node *tree = new Node[n];
for (int i=0; iint tmp, left, right;
in >> tmp >> left >> right;
tree[i].data = tmp;
if (left != -1) tree[i].left = &tree[left];
else tree[i].left = NULL;
if (right != -1) tree[i].right = &tree[right];
else tree[i].right = NULL;
}
TreePrint tp;
tp.bfs(&tree[0]);
tp.display();
in >> n;
}
in.close();
return 0;
}
更简单的方法
由于输入的时候就是BFS遍历，所以输入的时候就可以进行排序
#include
#include
#include
#include
using namespace std;
struct Node {
int data;
int id;
int colum;
};
bool compareTo(Node &l, Node &r) {
if (l.colum == r.colum) return l.id < r.id;
else return l.colum < r.colum;
}
int main() {
ifstream in("../input.txt");
int n;
in >> n;
while (n != -1) {
queue q;
q.push(0);
Node *tree = new Node[n];
tree[0].colum = 0;
for (int i=0; iint tmp, left, right;
in >> tmp >> left >> right;
tree[i].data = tmp;
tree[i].id = i;
int pos = q.front();
q.pop();
if (left != -1) {
tree[left].colum = pos - 1;
q.push(pos-1);
}
if (right != -1) {
tree[right].colum = pos + 1;
q.push(pos+1);
}
}
sort(tree, tree+n, &compareTo);
for (int i=0; iprintf("%d ", tree[i].data);
}
printf("\n");
in >> n;
}
in.close();
return 0;
}