题目
在二叉树中找到一个节点的后继节点
【题目】 现在有一种新的二叉树节点类型如下:
public class Node { public int value; public Node left;
public Node right; public Node parent;
public Node(int data) { this.value = data; }
} 该
结构比普通二叉树节点结构多了一个指向父节点的parent指针。 假
设有一 棵Node类型的节点组成的二叉树, 树中每个节点的parent指针
都正确地指向 自己的父节点, 头节点的parent指向null。 只给一个在
二叉树中的某个节点 node, 请实现返回node的后继节点的函数。 在二
叉树的中序遍历的序列中, node的下一个节点叫作node的后继节点。
思路
记住这两条规律:
1.若当前结点有右子树时,则一直找它的最左端结点,直到没有左结点,返回该结点就是后继结点;
2,若·当前结点没有右子树时,则判断当前结点的父结点的左结点是否是当前结点,如果是则返回父节点,如果不是则令父节点为·当前节点,依次重复上述过程,直到找到为止,
代码
// 中序遍历寻找后继结点->cpp : 定义控制台应用程序的入口点。
//
#include "stdafx.h"
#include<iostream>
using namespace std;
struct Node
{
int val;
struct Node*parent;
struct Node*left;
struct Node*right;
struct Node(int data) : val(data), left(nullptr), right(nullptr) {};
};
Node*getmostleft(Node*node);
Node* getsucessful(Node*node)
{
if (node == NULL)
return node;
if (node->right != NULL)
{
return getmostleft(node->right);
}
else {
Node*parent = node->parent;
while (parent != NULL&&parent->left != node)
{
node = parent;
parent = node->parent;
}
return parent;
}
}
Node*getmostleft(Node*node)
{
if (node == NULL)
return node;
while (node->left != NULL)
{
node = node->left;
}
return node;
}
int main()
{
Node*head= new Node(10);
head->parent = NULL;
head->left = new Node(3);
head->left->parent = head;
head->left->left = new Node(1);
head->left->left->parent = head->left;
head->left->left->right = new Node(2);
head->left->left->right->parent = head->left->left;
head->left->right = new Node(4);
head->left->right->parent = head->left;
head->left->right->right = new Node(5);
head->left->right->right->parent = head->left->right;
head->right = new Node(9);
head->right->parent = head;
head->right->left = new Node(8);
head->right->left->parent = head->right;
head->right->left->left = new Node(7);
head->right->left->left->parent = head->right->left;
head->right->right = new Node(10);
head->right->right->parent = head->right;
Node* test = head->left->left;
Node*result1 = getsucessful(test);
cout << result1->val << " " << endl;
Node*test1 = head->left->left->right;
Node*result2= getsucessful(test1);
cout << result2->val << " " << endl;
Node*test02 = head->right->right;
Node*result3 = getsucessful(test02);
cout << result3<< " " << endl;
Node*test04= test = head->right->left->left;
Node*result4 = getsucessful(test04);
cout << result4->val << " " << endl;
return 0;
}