给定一颗二叉树和其中一个结点,如何找出中序遍历序列的下一个结点?
本文中有关二叉树结点结构体以及其它相关函数接口定义如下:
#include<stdio.h>
#include<stdlib.h>
#include<assert.h>
typedef int DataType;
typedef struct BSTreeNode{
DataType data;
struct BSTreeNode *left;
struct BSTreeNode *right;
} BSTreeNode;
BSTreeNode *CreateNode(int data)
{
BSTreeNode *node = (BSTreeNode *)malloc(sizeof(BSTreeNode));
node->data = data;
node->left = NULL;
node->right = NULL;
return node;
}
首先利用栈实现二叉树的非递归中序遍历,从根结点开始,当根结点不为空或者栈不为空的条件下,开始遍历左子树。每遍历一个结点,当前结点就入栈(结点为NULL不入栈)。当遍历到叶子结点时返回到叶子结点的父节点,开始遍历右子树(即弹出栈顶元素,此时栈顶的元素为叶子结点的父节点,此时遍历父节点的右子树)。
当在遍历过程中找到给定的结点时,标记此节点,以此节点为当前结点遍历二叉树,找到此节点在二叉树中序遍历序列中的下一个结点时,打印此结点data值即可。
有关栈的定义如下:
#include<assert.h>
struct BSTreeNode;
typedef struct BSTreeNode * QDataType;
#define MAXSIZE (100)
typedef struct{
QDataType array[MAXSIZE];
int top;
}Stack;
void StackInit(Stack *ps)
{
ps->top = 0;
}
void StackDestory(Stack *ps)
{
ps->top = 0;
}
void StackPush(Stack *ps, QDataType data)
{
assert(ps->top < MAXSIZE);
ps->array[ps->top++] = data;
}
void StackPop(Stack *ps)
{
assert(ps->top>0);
ps->top--;
}
QDataType StackTop(const Stack *ps)
{
assert(ps->top > 0);
return ps->array[ps->top - 1];
}
int StackSize(const Stack *ps)
{
return ps->top;
}
int StackEmpty(const Stack *ps)
{
return ps->top == 0 ? 1 : 0;
}
代码如下:
void InOrderLoopFind(BSTreeNode *root,DataType data1)
{
Stack stack;
StackInit(&stack);
BSTreeNode *cur, *top,*cur1;
cur = root;
while (cur != NULL || !StackEmpty(&stack))
{
//遍历左子树
while (cur != NULL)
{
StackPush(&stack, cur);
cur = cur->left;
}
//top的左子树已经遍历
top = StackTop(&stack);
StackPop(&stack);
//在树中遍历到给定的那个结点
if (top->data == data1)
{
cur1 = top->right;
while (cur1 != NULL || !StackEmpty(&stack))
{
//遍历左子树
while (cur1 != NULL)
{
StackPush(&stack, cur1);
cur1 = cur1->left;
}
//top的左子树已经遍历
top = StackTop(&stack);
StackPop(&stack);
//打印当前结点
printf("%d", top->data);
}
}
//利用子问题的思想遍历top右子树
cur = top->right;
}
}