追寻伯利恒二叉树 -- 不用栈不用递归

#include<stdio.h>
#include<stdlib.h>

struct tNode
{
   int data;
   struct tNode* left;
   struct tNode* right;
};

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
    {
      /* 找到current的前驱节点 */
      pre = current->left;
      while(pre->right != NULL && pre->right != current)
        pre = pre->right;

      /* 将current节点作为其前驱节点的右孩子 */
      if(pre->right == NULL)
      {
        pre->right = current;
        current = current->left;
      }

      /* 恢复树的原有结构，更改right 指针 */   
      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 */
}

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);
}

/* 测试*/
int main()
{

  /* 构建树结构如下：
            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);
   return 0;
}