根據morris inorder，這棵樹的下一步是什麼？

Question

就在我坐下來寫代碼莫里斯序遍歷我想這個例子，我有點困惑就如何在這種特殊情況下，它會工作：

 80 
    / \ 
    60  100 
    \  /
    70 90 
    /
    65 
/
63 
    \ 
    64 

第1步：

60 
    \ 
    70 
/ \ 
    65  80 
/  \ 
63  100 
\  /
    64  90 

據我所知，下一步70中的算法將成爲65的正確孩子，那麼60會發生什麼？我很確定我錯過了一些微不足道的東西，但不幸的是我無法將它放在手指上。

public void MorrisInorder() { 
    BSTNode<T> p = root, tmp; 
    while (p != null) 
     if (p.left == null) { 
      visit(p); 
      p = p.right; 
     } 
     else { 
      tmp = p.left; 
      while (tmp.right != null && // go to the rightmost node of 
        tmp.right != p) // the left subtree or 
        tmp = tmp.right; // to the temporary parent of p; 
      if (tmp.right == null) {// if 'true' rightmost node was 
        tmp.right = p;  // reached, make it a temporary 
        p = p.left;  // parent of the current root, 
      } 
      else {     // else a temporary parent has been 
        visit(p);   // found; visit node p and then cut 
        tmp.right = null; // the right pointer of the current 
        p = p.right;  // parent, whereby it ceases to be 
      }      // a parent; 
     } 
} 

代碼我正在關注morris inorder遍歷。

Answer 1

直接回答你的問題，我認爲這個數字是不準確，你的情況的第1步，從節點「80」邊緣節點「60」不應該被刪除。步驟1中唯一的改變是將節點「70」的右點重定向到節點「80」（參見步驟1），這指示算法經過節點「80」的左子樹之後的返回路徑。

步驟1：

 80 
    /^ \ 
    60 | 100 
    \ | /
    70 90 
    /
    65 
/
63 
    \ 
    64 

添加來自節點「70」返回路徑到節點「80」，作爲當前節點「60」的左側點之後爲NULL，則當前節點將被設置作爲節點「70」。同時，節點「65」的右側點將被重定向到節點「70」

步驟2：

 80 
    /^ \ 
    60 | 100 
    \ | /
    70 90 
    /^ 
/| 
    65 
/
63 
    \ 
    64 

詳情莫里斯的序遍歷的代碼被列爲如下。

假設我們有一個節點結構等：

/* 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; 
}; 

和遍歷是：

/* 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) 
    { 
    /* This means there is no left sub-tree for current node, 
     then just print current node, and go to the right "child" node. 
     The right "child" node may be either its true child node, 
     or the returning path for "60" sub-tree (like "70" to "80") */ 
    if(current->left == NULL) 
    { 
     printf(" %d ", current->data); 
     current = current->right;  
    }  
    else 
    { 
     /* before going to the left sub-tree, we need to find a returning path 
     to current node (such as when current node is "80", and we want to 
     go to "60", so we need to save the returning path from left sub-tree 
     to "80"). It is easy to imagine that we need to return to the current 
     node when we arriving the right-most node of current left sub-tree. 
     Therefore, we just go to the right-most node (the first condition in 
     while) and set the returning path at "pre->right == NULL" block, as 
     well as updating the current node. Another situation is that when we 
     arrive at the left-most leaf node (if not exist, it means current->left 
     is NULL, and we won't go into this block), we have already set the right 
     point of left-most leaf node as the returning node (it un-satisfies the 
     second condition of while loop), and then we will recover the right 
     point of this leaf node in the next "else" block. 
      */ 
     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 `enter code here` 
     { 
     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 */ 
}