Morris Traversal算法非递归法遍历二叉树

网上找了一个中文的ppt，结果里面写的是错的，坑爹！！顺便提一下，Morris就是KMP算法里的那个“M”。算法的时间复杂度是O(n)，没有使用额外的空间O(1)。

这个算法主要是利用了线索二叉树，在遍历过程中，将树变形了，最后又还原了。以下是转载：

Using Morris Traversal, we can traverse the tree without using stack and recursion. The idea of Morris Traversal is based on Threaded Binary Tree. In this traversal, we first create links to Inorder successor and print the data using these links, and finally revert the changes to restore original tree. 

1. Initialize current as root2. While current is not NULL   If current does not have left child      a) Print current’s data      b) Go to the right, i.e., current = current->right   Else      a) Make current as right child of the rightmost node in current's left subtree      b) Go to this left child, i.e., current = current->left

Although the tree is modified through the traversal, it is reverted back to its original shape after the completion. Unlike Stack based traversal, no extra space is required for this traversal.

?
#include<stdio.h>
#include<stdlib.h>
/* A binary tree tNode has data, pointer to left child
   and a pointer to right child */
struct tNode
{
   int data;
   struct tNode* left;
   struct tNode* right;
};
/* Function to traverse binary tree without recursion and
   without stack */
void MorrisTraversal(struct tNode *root)
{
  struct tNode *current,*pre;
  if(root == NULL)
     return;
  current = root;
  while(current != NULL)
  {
    if(current->left == NULL)
    {
      printf(" %d ", current->data);
      current = current->right;
    }
    else
    {
      /* Find the inorder predecessor of current */
      pre = current->left;
      while(pre->right != NULL && pre->right != current)
        pre = pre->right;
      /* Make current as right child of its inorder predecessor */
      if(pre->right == NULL)
      {
        pre->right = current;
        current = current->left;
      }
      /* Revert the changes made in if part to restore the original
        tree i.e., fix the right child of predecssor */
      else
      {
        pre->right = NULL;
        printf(" %d ",current->data);
        current = current->right;
      } /* End of if condition pre->right == NULL */
    } /* End of if condition current->left == NULL*/
  } /* End of while */
}
/* UTILITY FUNCTIONS */
/* Helper function that allocates a new tNode with the
   given data and NULL left and right pointers. */
struct tNode* newtNode(int data)
{
  struct tNode* tNode = (struct tNode*)
                       malloc(sizeof(struct tNode));
  tNode->data = data;
  tNode->left = NULL;
  tNode->right = NULL;
  return(tNode);
}
/* Driver program to test above functions*/
int main()
{
  /* Constructed binary tree is
            1
          /   \
        2      3
      /  \
    4     5
  */
  struct tNode *root = newtNode(1);
  root->left        = newtNode(2);
  root->right       = newtNode(3);
  root->left->left  = newtNode(4);
  root->left->right = newtNode(5);
  MorrisTraversal(root);
  getchar();
  return 0;
}

References:
www.liacs.nl/~deutz/DS/september28.pdf
http://comsci.liu.edu/~murali/algo/Morris.htm
www.scss.tcd.ie/disciplines/software_systems/…/HughGibbonsSlides.pdf
http://www.mec.ac.in/resources/notes/notes/ds/thread.htm


PS：还是老外写的简洁易懂，赞~

原文链接：http://geeksforgeeks.org/archives/6358

 
posted @ 2011-09-03 13:42  dc0453  阅读(456)  评论(0)  编辑  收藏  举报