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//二叉树常用操作
//Thu Jul 17 06:40:24 PDT 2014
#include <iostream>
#include <assert.h>
#include <stack>
#include <queue>
#include <vector>
using namespace std;
typedef struct BTreeNode {
int value;
BTreeNode *lchild;
BTreeNode *rchild;
}*BTree;
//函数：根据所给数据创建二叉排序树
BTree CreateBinaryTree(int *arr, int n);
//向二叉排序树中插入元素-非递归
void InsertBinaryTree(BTree *root, int val);
//函数：二叉树遍历-前序遍历
void PreOrderTraversal(BTree root);
//函数：二叉树遍历-中序遍历
void InOrderTraversal(BTree root);
//函数：二叉树遍历-后序遍历
void PostOrderTraversal(BTree root);
//函数：重建二叉树，依据前序遍历结果和中序遍历结果
BTreeNode* BuildBTreeWithInorderPreOrder(int pre[], int in[], int n);
BTreeNode* Construct(int *preStart, int *preEnd, int *inStart, int *inEnd);
//函数：判断二叉树的后序遍历是否合法
bool IsPostOrderValid(int post[], int n);
//函数：二叉树中和为某值的路径
void FindPathWithSum(BTree root, int expectedSum);
void FindPathCore(BTree root, vector<BTreeNode*> &path, int currentSum, int expectedSum);
//函数：将二叉树转化成双向链表
BTreeNode* ConvertNode(BTree root);
void ConvertNodeCore(BTreeNode* curNode, BTreeNode* &preNode);
//函数：求二叉树深度
int TreeDepth(BTree root);
//函数：判断二叉树是否平衡二叉树
bool IsAVLTree(BTree root);
//函数：层次遍历二叉树
void TraversalByLevel(BTree root);
//求二叉树第K层节点个数
int GetNodeNumKthLevel(BTree root, int k);
//求二叉树中两个节点的最大距离
int GetMaxDistance(BTree root);
//判断一颗二叉树是否为完全二叉树
bool IsCompleteTree(BTree root);
//求一个二叉树中两个节点最低公共祖先节点
BTreeNode* GetLCANode(BTree root, BTreeNode* u, BTreeNode *v);
bool IsNodeExist(BTree root, BTreeNode* node);
int main()
{
//int a[] = {7, 2, 8, 3, 6, 4, 5, 1};
//int n = sizeof(a) / sizeof(a[0]);
//BTree bt = CreateBinaryTree(a, n);
//cout << "前序遍历：";
//PreOrderTraversal(bt);
//cout << endl;
//cout << "中序遍历：";
//InOrderTraversal(bt);
//cout << endl;
//cout << "后序遍历：";
//PostOrderTraversal(bt);
//cout << endl;
//cout << "层次遍历：";
//TraversalByLevel(bt);
//cout << endl;
int preOrder[] = { 7, 2, 1, 3, 6, 4, 5, 8 };
int inOrder[] = { 1, 2, 3, 4, 5, 6, 7, 8 };
int n = sizeof(preOrder) / sizeof(int);
BTree bt = BuildBTreeWithInorderPreOrder(preOrder, inOrder, n);
cout << "前序遍历：" ;
PreOrderTraversal(bt);
cout << endl;
cout << "中序遍历：" ;
InOrderTraversal(bt);
cout << endl;
cout << "后序遍历：" ;
PostOrderTraversal(bt);
cout << endl;
int postOrder[] = { 1, 5, 4, 6, 3, 2, 8, 7};
cout << "后续序列合法：" << IsPostOrderValid(postOrder, n)
<< "\t树深度：" << TreeDepth(bt) << endl;
FindPathWithSum(bt, 27);
cout << "AVL：" << IsAVLTree(bt)
<< "\t第3层节点数：" << GetNodeNumKthLevel(bt, 3) << endl;
cout << "是否完全二叉树：" << IsCompleteTree(bt) << endl;
cout << "两个节点最大距离：" << GetMaxDistance(bt) << endl;
BTreeNode* u = bt->lchild->lchild;
BTreeNode* v = bt->lchild->rchild->rchild->lchild;
cout << IsNodeExist(bt, u) << "|" << IsNodeExist(bt, v) << "|" << IsNodeExist(bt, NULL) << endl ;
cout << u->value << " " << v->value << endl;
cout << (GetLCANode(bt, u, v))->value << endl; //有BUG，待完善
//BTreeNode* head = ConvertNode(bt);
//BTreeNode* pos = head;
//cout << "转换成的链表遍历：";
//while (head != NULL)
//{
// cout << head->value << "\t";
// pos = head->rchild;
//}
//cout << endl;
return 0;
}
//函数：根据所给数据创建二叉排序树
BTree CreateBinaryTree(int *arr, int n)
{
assert(arr != NULL && n > 0);
BTree root = NULL;
for (int i=0; i<n; ++i)
InsertBinaryTree(&root, *(arr+i));
//InsertBTreeRecursion(root, *(arr+i));
return root;
}
//向二叉排序树中插入元素-非递归
void InsertBinaryTree(BTree *root, int val)
{
BTreeNode *newNode = new BTreeNode();
newNode->value = val;
newNode->lchild = newNode->rchild = NULL;
BTreeNode *pre, *cur;
pre = NULL;
cur = *root;
while (cur != NULL)
{
pre = cur;
if (val < cur->value)
cur = cur->lchild;
else
cur = cur->rchild;
}
if (NULL == pre)
*root = newNode;
else if (val < pre->value)
pre->lchild = newNode;
else
pre->rchild = newNode;
}
//函数：二叉树遍历-前序遍历
void PreOrderTraversal(BTree root)
{
if (root != NULL)
{
cout << root->value << "\t";
PreOrderTraversal(root->lchild);
PreOrderTraversal(root->rchild);
}
}
//函数：二叉树遍历-中序遍历
void InOrderTraversal(BTree root)
{
if (root != NULL)
{
InOrderTraversal(root->lchild);
cout << root->value << "\t";
InOrderTraversal(root->rchild);
}
}
//函数：二叉树遍历-后序遍历
void PostOrderTraversal(BTree root)
{
if (root != NULL)
{
PostOrderTraversal(root->lchild);
PostOrderTraversal(root->rchild);
cout << root->value << "\t";
}
}
//函数：重建二叉树，依据前序遍历结果和中序遍历结果
//思路：1.从前序系列中取第一个元素x，再在中序序列中找到这个元素x
// 2.以x为界，以x的值新建根节点root，x左边即为左子树(记节点数为ln),x右边即为右子树(记节点数为rn)
// 3.在前序序列中x后，前ln个节点即为左子树，后rn个节点即为右子树
// 4.设置root的左右孩子分别为1、2、3步递归调用左右子树序列后的返回值
BTreeNode* BuildBTreeWithInorderPreOrder(int pre[], int in[], int n)
{
assert(pre != NULL && in != NULL && n > 0);
return Construct(pre, pre + n - 1, in, in + n - 1);
}
BTreeNode* Construct(int *preStart, int *preEnd, int *inStart, int *inEnd)
{
int rootVal = *preStart;
BTreeNode *root = new BTreeNode();
root->value = rootVal;
root->lchild = root->rchild = NULL;
//定义递归出口：当前序、中序序列中都仅有一个元素时，递归返回该节点
if (preStart == preEnd)
{
if (inStart == inEnd && *preStart == *inStart)
return root;
else
throw "Invalid Input.";
}
//在中序序列中查找根节点
int *pos = inStart;
while (pos <= inEnd && *pos != rootVal)
++pos;
if (pos == inEnd && *pos != rootVal)
throw "Invalid Input.";
int leftLength = pos - inStart;
int *leftPreEnd = preStart + leftLength;
//构建左子树
if (leftLength > 0)
{
root->lchild = Construct(preStart + 1, leftPreEnd, inStart, pos - 1);
}
//构建右子树
if (leftLength < preEnd - preStart)
{
root->rchild = Construct(leftPreEnd + 1, preEnd, pos + 1, inEnd);
}
return root;
}
//函数：判断二叉树的后序遍历是否合法
bool IsPostOrderValid(int post[], int n)
{
assert(post != NULL && n > 0);
//取序列最后一个元素作为根
int root = post[n-1];
//扫描序列，比根小作为左子树
int leftLength, rightLength;
int *lend = post;
while(lend <= post + n - 1 && *lend < root)
++lend;
leftLength = lend - post;
//比根大作为右子树
int *rend = lend;
while(rend <= post + n - 1)
{
if (*rend < root)
return false;
++rend;
}
rightLength = n - leftLength - 1;
//判断左子树是否二叉搜索树
bool left = true;
if (leftLength > 0)
left = IsPostOrderValid(post, leftLength);
//判断左子树是否二叉搜索树
bool right = true;
if (rightLength > 0)
left = IsPostOrderValid(rend, rightLength);
return left && right;
}
//函数：二叉树中和为某值的路径
void FindPathWithSum(BTree root, int expectedSum)
{
assert(root != NULL);
vector<BTreeNode*> path;
FindPathCore(root, path, 0, expectedSum);
}
void FindPathCore(BTree root, vector<BTreeNode*> &path,
int currentSum, int expectedSum)
{
currentSum += root->value;
path.push_back(root);
//递归出口：如果路径到达页节点，且和为期望值，则打印
bool isLeaf = (root->lchild == NULL && root->rchild == NULL);
if (isLeaf && currentSum == expectedSum)
{
vector<BTreeNode*>::iterator ite = path.begin();
for (;ite != path.end(); ++ite)
cout << (*ite)->value << "\t";
cout << endl;
}
//如果不是叶节点，则遍历他的子节点
if (root->lchild != NULL)
FindPathCore(root->lchild, path, currentSum, expectedSum);
if (root->rchild != NULL)
FindPathCore(root->rchild, path, currentSum, expectedSum);
//该条路径不合要求，路径上删除该节点
//currentSum -= root->value; //不需要，因为非引用传递
path.pop_back();
}
//函数：将二叉树转化成双向链表
BTreeNode* ConvertNode(BTree root)
{
assert(root != NULL);
BTreeNode* lastNode = NULL;
ConvertNodeCore(root, lastNode);
BTreeNode* head = lastNode;
while(head != NULL && head->lchild != NULL)
head = head->lchild;
return head;
}
void ConvertNodeCore(BTreeNode* curNode, BTreeNode* &preNode)
{
//可以借用中序遍历的思路
if (curNode == NULL)
return;
if(curNode->lchild != NULL)
ConvertNodeCore(curNode->lchild, preNode);
//curNode前驱指向preNode, preNode后继指向curNode
curNode->lchild = preNode;
if (preNode != NULL)
preNode->rchild = curNode;
preNode = curNode;
if(curNode->rchild != NULL)
ConvertNodeCore(curNode->rchild, preNode);
}
//函数：求二叉树深度
int TreeDepth(BTree root)
{
if (root == NULL)
return 0;
int llen = TreeDepth(root->lchild);
int rlen = TreeDepth(root->rchild);
return 1 + (llen > rlen ? llen : rlen);
}
//函数：判断二叉树是否平衡二叉树
bool IsAVLTree(BTree root)
{
if (root == NULL)
return true;
int llen = TreeDepth(root->lchild);
int rlen = TreeDepth(root->rchild);
if (llen - rlen <= -1 || llen - rlen >= 1)
return false;
else
return IsAVLTree(root->lchild) && IsAVLTree(root->rchild);
}
//层次遍历二叉树
void TraversalByLevel(BTree root)
{
assert(root != NULL);
queue<BTreeNode*> que;
que.push(root);
while(!que.empty())
{
BTreeNode *node = que.front();
cout << node->value << "\t";
que.pop();
if (node->lchild != NULL)
que.push(node->lchild);
if (node->rchild != NULL)
que.push(node->rchild);
}
}
//求二叉树第K层节点个数
int GetNodeNumKthLevel(BTree root, int k)
{
if (root == NULL)
return 0;
if (k == 1)
return 1;
return GetNodeNumKthLevel(root->lchild, k - 1)
+ GetNodeNumKthLevel(root->rchild, k - 1);
}
//求二叉树中两个节点的最大距离
int GetMaxDistance(BTree root)
{
return TreeDepth(root->lchild) + TreeDepth(root->rchild);
}
//判断一颗二叉树是否为完全二叉树
bool IsCompleteTree(BTree root)
{
//按照层次遍历，当遇到一个节点的左子树为空时，则该节点右子树必须为空。
//且后面遍历的节点必须都为叶节点
assert(root != NULL);
queue<BTreeNode*> que;
que.push(root);
bool leaf = false;
while (!que.empty())
{
BTreeNode* node = que.front();
que.pop();
if(leaf && (node->lchild != NULL || node->rchild != NULL)) //若应是叶节点却不是
return false;
if (node->lchild == NULL && node->rchild != NULL) //若左孩子为NULL，右却不为NULL
return false;
if (node->rchild == NULL)
leaf = true;
if (root->lchild != NULL)
que.push(root->lchild);
if (root->rchild != NULL)
que.push(root->rchild);
}
}
//求一个二叉树中两个节点最低公共祖先节点
BTreeNode* GetLCANode(BTree root, BTreeNode* u, BTreeNode *v)
{
if (root == NULL)
return NULL;
//设当前节点为root
//若当前节点左子树（或右子树）中能同时找到节点u和v，则将当前节点设置为root的左孩子（或右孩子）
//若节点u,v一个在当前节点的左子树，一个在当前节点的右子树，则当前节点为u，v最低公共祖先节点
BTreeNode *currentNode = root;
while (currentNode != NULL)
{
if (IsNodeExist(root->lchild, u) && IsNodeExist(root->rchild, v)
|| IsNodeExist(root->lchild, v) && IsNodeExist(root->rchild, u))
return currentNode;
if (IsNodeExist(root->lchild, u) && IsNodeExist(root->lchild, v))
currentNode = currentNode->lchild;
if (IsNodeExist(root->rchild, u) && IsNodeExist(root->rchild, v))
currentNode = currentNode->rchild;
}
return NULL;
}
bool IsNodeExist(BTree root, BTreeNode* node)
{
if (root != NULL)
{
if (root == node)
return true;
return IsNodeExist(root->lchild, node)
|| IsNodeExist(root->rchild, node);
return false;
}
return false;
}