Morris Traversal - 不需遞歸和棧中序遍歷二叉樹

Morris Traversal的目的是在不適用遞歸和棧的情況下，在相同時間內得到二叉樹中序遍歷結果，空間複雜度爲O(1)

基本思想是線索二叉樹：首先建立中序遍歷的後向鏈接指針，然後通過指針進行中序遍歷，遍歷後再將二叉樹還原。

1. Initialize current as root 
2. 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

注意：1、關鍵一步是倒數第二行，一個節點在中序遍歷的前趨節點是其左子樹的最右邊的節點（一直向右兒子遍歷直到其不存在）

2、遍歷完成後，若在倒數第二行這一步找到的節點爲current自己，那麼將前一個節點的右兒子置爲空，即還原二叉樹

#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;
}

參考：http://www.geeksforgeeks.org/inorder-tree-traversal-without-recursion-and-without-stack/

相關題目：https://oj.leetcode.com/problems/recover-binary-search-tree/