Question

我试图按顺序遍历二叉搜索树，并将数据（已排序）放入数组中。由于某种原因，指向数组中当前位置的指针没有向右移动。

这是DS的声明：

TreeRoot    DWORD              Null (left child)
            DWORD   Null (right child)
            SDWORD   6 (numeric value)

这是我正在尝试编写的函数：

TreeToArray PROC
    rootPtr=8;
    ArrayPtr=rootPtr+4;

    ;Saving the Registers 
    push ebp;
    mov ebp,esp;
    push esi;
    push edx;
    push ebx;
    push edi;
    push ecx;

Check:
    mov esi,rootPtr[ebp]; esi holds the current root
    mov edi, ArrayPtr[ebp] ;edi holds the pointer to the array
    cmp esi,Null ;if root=null
    je Done2;

LeftSubTree:
    push edi
    push BinTreeLeft[esi]
    call TreeToArray; recursive call for left sub tree

Visit:
    mov ebx,BinTreeValue[esi] ;getting the value of the node
    mov [edi],ebx
    add edi,4

RightSubTree:
    push edi
    push BinTreeRight[esi]
    call TreeToArray; recursive call for right sub tree

Done2:
    pop ecx;
    pop edi;
    pop ebx
    pop edx
    pop esi
    pop ebp
    ret 8;

TreeToArray ENDP

Answer 1

您的代码现在看起来像这样（如果用C拼写）：

typedef struct tNode
{
  struct tNode* pLeftChild;
  struct tNode* pRightChild;
  int Value;
} tNode;

void TreeToArray(tNode* pNode, int* pArrayElement)
{
  if (pNode == NULL) return;

  TreeToArray(pNode->pLeftChild, pArrayElement);

  *pArrayElement = pNode->Value;
  pArrayElement++;

  TreeToArray(pNode->pRightChild, pArrayElement);
}

只有在转到正确的子节点并且在返回父节点时忘记推进指针时，才“移动”指针。

您想要做的是：

int* TreeToArray(tNode* pNode, int* pArrayElement)
{
  if (pNode == NULL) return pArrayElement;

  pArrayElement = TreeToArray(pNode->pLeftChild, pArrayElement);

  *pArrayElement = pNode->Value;
  pArrayElement++;

  pArrayElement = TreeToArray(pNode->pRightChild, pArrayElement);

  return pArrayElement;
}

汇编：遍历二叉搜索树

1 个答案: