二叉树的逆时针螺旋遍历？

这里我们将看到一个有趣的问题。我们有一棵二叉树。我们必须以逆时针的方式遍历树。遍历的顺序如下所示 −

遍历序列是 1, 8, 9, 10, 11, 12, 13, 14, 15, 3, 2, 4, 5, 6, 7

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算法

antiClockTraverse(root)

Begin
   i := 1, j := height of the tree
   flag := false
   while i <= j, do
      if flag is false, then
         print tree elements from right to left for level i
         flag := true
         i := i + 1
      else
         print tree elements from left to right for level j
         flag := false
         j := j - 1
      end if
   done
End

示例

#include
using namespace std;
class Node {
   public:
      Node* left;
      Node* right;
      int data;
   Node(int data) { //constructor to create node
      this->data = data;
      this->left = NULL;
      this->right = NULL;
   }
};
int getHeight(Node* root) {
   if (root == NULL)
   return 0;
   //get height of left and right subtree
   int hl = getHeight(root->left);
   int hr = getHeight(root->right);
   return 1 + max(hl, hr); //add 1 for root
}
void printLeftToRight(class Node* root, int level) {
   if (root == NULL)
      return;
   if (level == 1)
      cout << root->data << " ";
   else if (level > 1) {
      printLeftToRight(root->left, level - 1);
      printLeftToRight(root->right, level - 1);
   }
}
void printRightToLeft(struct Node* root, int level) {
   if (root == NULL)
      return;
   if (level == 1)
      cout << root->data << " ";
   else if (level > 1) {
      printRightToLeft(root->right, level - 1);
      printRightToLeft(root->left, level - 1);
   }
}
void antiClockTraverse(class Node* root) {
   int i = 1;
   int j = getHeight(root);
   int flag = 0; //flag is used to change direction
   while (i <= j) {
      if (flag == 0) {
         printRightToLeft(root, i);
         flag = 1; //set flag to print from left to right
         i++;
      }else {
         printLeftToRight(root, j);
         flag = 0; //set flag to print from right to left
         j--;
      }
   }
}
int main() {
   struct Node* root;
   root = new Node(1);
   root->left = new Node(2);
   root->right = new Node(3);
   root->left->left = new Node(4);
   root->left->right = new Node(5);
   root->right->left = new Node(6);
   root->right->right = new Node(7);
   root->left->left->left = new Node(8);
   root->left->left->right = new Node(9);
   root->left->right->left = new Node(10);
   root->left->right->right = new Node(11);
   root->right->left->left = new Node(12);
   root->right->left->right = new Node(13);
   root->right->right->left = new Node(14);
   root->right->right->right = new Node(15);
   antiClockTraverse(root);
}

输出

1 8 9 10 11 12 13 14 15 3 2 4 5 6 7