解释不使用堆栈或递归的 Morris 中序树遍历

Question

有人可以帮助我理解以下莫里斯中序树遍历算法而不使用堆栈或递归吗？我试图理解它是如何工作的，但它只是逃避了我。

 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. In current's left subtree, make current the right child of the rightmost node
   b. Go to this left child, i.e., current = current->left

我了解树的修改方式是将当前节点作为右子树中最大节点的右子节点 并使用此属性进行中序遍历。但除此之外，我迷失了。

编辑： 找到了附带的 C++ 代码。我很难理解树被修改后如何恢复。神奇之处在于 else 子句，一旦右叶被修改，就会触发该子句。详情见代码：

/* 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;
      }

     // MAGIC OF RESTORING the Tree happens here: 
      /* 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 */
}

Answer 1

如果我正确地阅读了该算法，这应该是其工作原理的示例：

     X
   /   \
  Y     Z
 / \   / \
A   B C   D

首先，X 是根，因此它被初始化为current。 X 有一个左子树，因此 X 成为 X 左子树的最右边的右子树——的直接前身X 中序遍历。因此，X 成为 B 的右子节点，然后 current 设置为 Y。树现在看起来像这样：

    Y
   / \
  A   B
       \
        X
       / \
     (Y)  Z
         / \
        C   D

上面的 (Y) 指的是 Y 及其所有子节点，由于递归问题而被省略。无论如何，重要的部分都列出来了。
现在树有一个返回X的链接，遍历继续...

 A
  \
   Y
  / \
(A)  B
      \
       X
      / \
    (Y)  Z
        / \
       C   D

然后输出A，因为它没有左孩子，并且current返回到Y，在上一次迭代中成为 A 的右子节点。在下一次迭代中，Y 拥有两个孩子。然而，循环的双重条件使其在到达自身时停止，这表明它的左子树已被遍历。因此，它打印自身，并继续其右子树，即 B。

B 打印自身，然后 current 变为 X，它也经历与 Y 相同的检查过程意识到它的左子树已被遍历，继续Z。树的其余部分遵循相同的模式。

不需要递归，因为不是依赖于通过堆栈进行回溯，而是将返回到（子）树根的链接移动到无论如何都会在递归中序树遍历算法中访问它的点 - 在其结束之后左子树已经完成。

Answer 2

递归中序遍历为：(in-order(left)->key->in-order(right))。 （这与DFS类似）

当我们进行DFS时，我们需要知道回溯到哪里（这就是我们通常保留堆栈的原因）。

当我们遍历父节点时，我们需要回溯到 ->我们找到需要回溯的节点并将其链接更新到父节点。

当我们回溯时？当我们无法走得更远的时候。什么时候我们不能走得更远？当没有留下孩子的礼物时。

我们回溯到哪里？注意：致继承人！

因此，当我们沿着左子路径跟踪节点时，将每一步的前驱节点设置为指向当前节点。这样，前任就会有到后继者的链接（回溯的链接）。

我们尽可能向左行驶，直到需要原路返回。当我们需要回溯时，我们打印当前节点并沿着正确的链接到达后继节点。

如果我们刚刚回溯->我们需要跟随右孩子（我们已经完成了左孩子）。

如何判断我们是否刚刚回溯呢？获取当前节点的前驱节点并检查它是否有正确的链接（到该节点）。如果有——那么我们就遵循它。删除链接以恢复树。

如果没有左侧链接=>我们没有走回头路，应该继续追随剩下的孩子。

这是我的 Java 代码（抱歉，它不是 C++）

public static <T> List<T> traverse(Node<T> bstRoot) {
    Node<T> current = bstRoot;
    List<T> result = new ArrayList<>();
    Node<T> prev = null;
    while (current != null) {
        // 1. we backtracked here. follow the right link as we are done with left sub-tree (we do left, then right)
        if (weBacktrackedTo(current)) {
            assert prev != null;
            // 1.1 clean the backtracking link we created before
            prev.right = null;
            // 1.2 output this node's key (we backtrack from left -> we are finished with left sub-tree. we need to print this node and go to right sub-tree: inOrder(left)->key->inOrder(right)
            result.add(current.key);
            // 1.15 move to the right sub-tree (as we are done with left sub-tree).
            prev = current;
            current = current.right;
        }
        // 2. we are still tracking -> going deep in the left
        else {
            // 15. reached sink (the leftmost element in current subtree) and need to backtrack
            if (needToBacktrack(current)) {
                // 15.1 return the leftmost element as it's the current min
                result.add(current.key);
                // 15.2 backtrack:
                prev = current;
                current = current.right;
            }
            // 4. can go deeper -> go as deep as we can (this is like dfs!)
            else {
                // 4.1 set backtracking link for future use (this is one of parents)
                setBacktrackLinkTo(current);
                // 4.2 go deeper
                prev = current;
                current = current.left;
            }
        }
    }
    return result;
}

private static <T> void setBacktrackLinkTo(Node<T> current) {
    Node<T> predecessor = getPredecessor(current);
    if (predecessor == null) return;
    predecessor.right = current;
}

private static boolean needToBacktrack(Node current) {
    return current.left == null;
}

private static <T> boolean weBacktrackedTo(Node<T> current) {
    Node<T> predecessor = getPredecessor(current);
    if (predecessor == null) return false;
    return predecessor.right == current;
}

private static <T> Node<T> getPredecessor(Node<T> current) {
    // predecessor of current is the rightmost element in left sub-tree
    Node<T> result = current.left;
    if (result == null) return null;
    while(result.right != null
            // this check is for the case when we have already found the predecessor and set the successor of it to point to current (through right link)
            && result.right != current) {
        result = result.right;
    }
    return result;
}

Answer 3

我在这里为该算法制作了一个动画：
https://docs.google.com/presentation/d/11GWAeUN0ckP7yjHrQkIB0WT9ZUhDBSa -WR0VsPU38fg/edit?usp=sharing

这应该有助于理解。蓝色圆圈是光标，每张幻灯片都是外部 while 循环的迭代。

这是morris遍历的代码（我从极客那里复制并修改为极客）：

def MorrisTraversal(root):
    # Set cursor to root of binary tree
    cursor = root
    while cursor is not None:
        if cursor.left is None:
            print(cursor.value)
            cursor = cursor.right
        else:
            # Find the inorder predecessor of cursor
            pre = cursor.left
            while True:
                if pre.right is None:
                    pre.right = cursor
                    cursor = cursor.left
                    break
                if pre.right is cursor:
                    pre.right = None
                    cursor = cursor.right
                    break
                pre = pre.right
#And now for some tests. Try "pip3 install binarytree" to get the needed package which will visually display random binary trees
import binarytree as b
for _ in range(10):
    print()
    print("Example #",_)
    tree=b.tree()
    print(tree)
    MorrisTraversal(tree)

Answer 4

我找到了莫里斯遍历的很好的图解。

Answer 5

我希望下面的伪代码更具启发性：

node = root
while node != null
    if node.left == null
        visit the node
        node = node.right
    else
        let pred_node be the inorder predecessor of node
        if pred_node.right == null /* create threading in the binary tree */
            pred_node.right = node
            node = node.left
        else         /* remove threading from the binary tree */
            pred_node.right = null 
            visit the node
            node = node.right

参考问题中的 C++ 代码，内部 while 循环找到当前节点的有序前驱。在标准二叉树中，前驱的右子节点必须为空，而在线程版本中，右子节点必须指向当前节点。如果右子节点为空，则将其设置为当前节点，从而有效地创建线程，其中用作返回点，否则必须存储在堆栈上。如果右子树不为空，则算法确保恢复原始树，然后继续遍历右子树（在这种情况下，已知左子树已被访问）。

Answer 6

public static void morrisInOrder(Node root) {
        Node cur = root;
        Node pre;
        while (cur!=null){
            if (cur.left==null){
                System.out.println(cur.value);      
                cur = cur.right; // move to next right node
            }
            else {  // has a left subtree
                pre = cur.left;
                while (pre.right!=null){  // find rightmost
                    pre = pre.right;
                }
                pre.right = cur;  // put cur after the pre node
                Node temp = cur;  // store cur node
                cur = cur.left;  // move cur to the top of the new tree
                temp.left = null;   // original cur left be null, avoid infinite loops
            }        
        }
    }

我认为这段代码会更好理解，只需使用 null 来避免无限循环，不必使用其他魔法。它可以轻松修改为预购。

Answer 7

Python解决方案
时间复杂度：O(n)
空间复杂度：O(1)

优秀的莫里斯中序遍历解释

class Solution(object):
def inorderTraversal(self, current):
    soln = []
    while(current is not None):    #This Means we have reached Right Most Node i.e end of LDR traversal

        if(current.left is not None):  #If Left Exists traverse Left First
            pre = current.left   #Goal is to find the node which will be just before the current node i.e predecessor of current node, let's say current is D in LDR goal is to find L here
            while(pre.right is not None and pre.right != current ): #Find predecesor here
                pre = pre.right
            if(pre.right is None):  #In this case predecessor is found , now link this predecessor to current so that there is a path and current is not lost
                pre.right = current
                current = current.left
            else:                   #This means we have traverse all nodes left to current so in LDR traversal of L is done
                soln.append(current.val) 
                pre.right = None       #Remove the link tree restored to original here 
                current = current.right
        else:               #In LDR  LD traversal is done move to R  
            soln.append(current.val)
            current = current.right

    return soln

Answer 8

Morris 中序遍历的 PFB 解释。

  public class TreeNode
    {
        public int val;
        public TreeNode left;
        public TreeNode right;

        public TreeNode(int val = 0, TreeNode left = null, TreeNode right = null)
        {
            this.val = val;
            this.left = left;
            this.right = right;
        }
    }

    class MorrisTraversal
    {
        public static IList<int> InOrderTraversal(TreeNode root)
        {
            IList<int> list = new List<int>();
            var current = root;
            while (current != null)
            {
                //When there exist no left subtree
                if (current.left == null)
                {
                    list.Add(current.val);
                    current = current.right;
                }
                else
                {
                    //Get Inorder Predecessor
                    //In Order Predecessor is the node which will be printed before
                    //the current node when the tree is printed in inorder.
                    //Example:- {1,2,3,4} is inorder of the tree so inorder predecessor of 2 is node having value 1
                    var inOrderPredecessorNode = GetInorderPredecessor(current);
                    //If the current Predeccessor right is the current node it means is already printed.
                    //So we need to break the thread.
                    if (inOrderPredecessorNode.right != current)
                    {
                        inOrderPredecessorNode.right = null;
                        list.Add(current.val);
                        current = current.right;
                    }//Creating thread of the current node with in order predecessor.
                    else
                    {
                        inOrderPredecessorNode.right = current;
                        current = current.left;
                    }
                }
            }

            return list;
        }

        private static TreeNode GetInorderPredecessor(TreeNode current)
        {
            var inOrderPredecessorNode = current.left;
            //Finding Extreme right node of the left subtree
            //inOrderPredecessorNode.right != current check is added to detect loop
            while (inOrderPredecessorNode.right != null && inOrderPredecessorNode.right != current)
            {
                inOrderPredecessorNode = inOrderPredecessorNode.right;
            }

            return inOrderPredecessorNode;
        }
    }