螺旋形式的层序遍历|使用双端队列
给定一棵二叉树，任务是打印给定树的螺旋顺序遍历。对于下面的树，函数应该打印 1、2、3、4、5、6、7。
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例子：
Input:
1
/ \
3 2
Output :
1
3 2
Input :
10
/ \
20 30
/ \
40 60
Output :
10
20 30
60 40我们已经看到使用两个堆栈的递归和迭代解决方案以及使用一个堆栈和一个队列的方法。在这篇文章中，讨论了一个带有一个双端队列的解决方案。这个想法是使用一个方向变量，并根据这个方向变量的值决定是从前面还是从后面弹出元素。
下面是上述方法的实现：
C++
// C++ program to print level order traversal
// in spiral form using one deque.
#include
using namespace std;
class Node {
public:
int data;
Node *left, *right;
Node(int val)
{
data = val;
left = NULL;
right = NULL;
}
};
void spiralOrder(Node* root)
{
deque d;
// Push root
d.push_back(root);
// Direction 0 shows print right to left
// and for Direction 1 left to right
int dir = 0;
while (!d.empty()) {
int size = d.size();
while (size--) {
// One whole level
// will be print in this loop
if (dir == 0) {
Node* temp = d.back();
d.pop_back();
if (temp->right)
d.push_front(temp->right);
if (temp->left)
d.push_front(temp->left);
cout << temp->data << " ";
}
else {
Node* temp = d.front();
d.pop_front();
if (temp->left)
d.push_back(temp->left);
if (temp->right)
d.push_back(temp->right);
cout << temp->data << " ";
}
}
cout << endl;
// Direction change
dir = 1 - dir;
}
}
int main()
{
// Build the Tree
Node* root = new Node(10);
root->left = new Node(20);
root->right = new Node(30);
root->left->left = new Node(40);
root->left->right = new Node(60);
// Call the Function
spiralOrder(root);
return 0;
} Java
// Java program to print level order traversal
// in spiral form using one deque.
import java.util.*;
class GFG
{
static class Node
{
int data;
Node left, right;
Node(int val)
{
data = val;
left = null;
right = null;
}
};
static void spiralOrder(Node root)
{
Deque d = new LinkedList();
// Push root
d.addLast(root);
// Direction 0 shows print right to left
// and for Direction 1 left to right
int dir = 0;
while (d.size() > 0)
{
int size = d.size();
while (size-->0)
{
// One whole level
// will be print in this loop
if (dir == 0)
{
Node temp = d.peekLast();
d.pollLast();
if (temp.right != null)
d.addFirst(temp.right);
if (temp.left != null)
d.addFirst(temp.left);
System.out.print(temp.data + " ");
}
else
{
Node temp = d.peekFirst();
d.pollFirst();
if (temp.left != null)
d.addLast(temp.left);
if (temp.right != null)
d.addLast(temp.right);
System.out.print(temp.data + " ");
}
}
System.out.println();
// Direction change
dir = 1 - dir;
}
}
// Driver code
public static void main(String args[])
{
// Build the Tree
Node root = new Node(10);
root.left = new Node(20);
root.right = new Node(30);
root.left.left = new Node(40);
root.left.right = new Node(60);
// Call the Function
spiralOrder(root);
}
}
// This code is contributed by Arnab Kundu Python3
# Python program to print level order traversal
# in spiral form using one deque.
class Node :
def __init__(self,val) :
self.data = val;
self.left = None;
self.right = None;
def spiralOrder(root) :
d = [];
# Push root
d.append(root);
# Direction 0 shows print right to left
# and for Direction 1 left to right
direct = 0;
while (len(d) != 0) :
size = len(d);
while (size) :
size -= 1;
# One whole level
# will be print in this loop
if (direct == 0) :
temp = d.pop();
if (temp.right) :
d.insert(0, temp.right);
if (temp.left) :
d.insert(0, temp.left);
print(temp.data, end= " ");
else :
temp = d[0];
d.pop(0);
if (temp.left) :
d.append(temp.left);
if (temp.right) :
d.append(temp.right);
print(temp.data ,end= " ");
print()
# Direction change
direct = 1 - direct;
if __name__ == "__main__" :
# Build the Tree
root = Node(10);
root.left = Node(20);
root.right = Node(30);
root.left.left = Node(40);
root.left.right = Node(60);
# Call the Function
spiralOrder(root);
# This code is contributed by AnkitRai01Javascript
输出：
10
20 30
60 40时间复杂度： O(N)
空间复杂度： O(N)
其中 N 是节点数