【二叉查找树】02不同的二叉查找树个数II【Unique Binary Search Trees II】

提到二叉查找树，就得想到二叉查找树的递归定义，

左子树的节点值都小于根节点，右子树的节点值都大于根节点。

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

给定一个n，问有多少个不同的二叉查找树，使得每个节点的值为 1...n?

例如，

给定n=3，你的程序应该返回所有的这5个不同的二叉排序树的个数。

   1         3     3      2      1

    \       /     /      / \      \

     3     2     1      1   3      2

    /     /       \                 \

   2     1         2                 3

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.

For example,
Given n = 3, your program should return all 5 unique BST's shown below.

   1         3     3      2      1

    \       /     /      / \      \

     3     2     1      1   3      2

    /     /       \                 \

   2     1         2                 3

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

test.cpp：

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#include <iostream>
#include <cstdio>
#include <stack>
#include <vector>
#include "BinaryTree.h"

using namespace std;

/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
*  int val;
*  TreeLinkNode *left, *right, *next;
*  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
vector<TreeNode *> generate(int start, int end)
{
    vector<TreeNode *> subTree;
    if (start > end)
    {
        subTree.push_back(NULL);
        return subTree;
    }

//对每个节点做root节点做遍历判断，当某个节点为root时候满足条件的二叉树可能有多个
    for (int k = start; k <= end; ++k)
    {
        vector<TreeNode *> leftSubTree = generate(start, k - 1);
        vector<TreeNode *> rightSubTree = generate(k + 1, end);
        for (int i = 0; i < leftSubTree.size(); ++i)
        {
            for (int j = 0; j < rightSubTree.size(); ++j)
            {
                TreeNode *tmp = new TreeNode(k);
                tmp->left = leftSubTree[i];
                tmp->right = rightSubTree[j];
                subTree.push_back(tmp);
            }
        }
    }
    return subTree;
}

vector<TreeNode *> generateTrees(int n)
{
    if (n == 0)
    {
        return generate(1, 0);
    }
    return generate(1, n);
}

vector<vector<int> > levelOrder(TreeNode *root)
{

vector<vector<int> > matrix;
    if(root == NULL)
    {
        return matrix;
    }
    vector<int> temp;
    temp.push_back(root->val);
    matrix.push_back(temp);

vector<TreeNode *> path;
path.push_back(root);

int count = 1;
    while(!path.empty())
    {
        TreeNode *tn = path.front();
        if(tn->left)
        {
            path.push_back(tn->left);
        }
        if(tn->right)
        {
            path.push_back(tn->right);
        }
        path.erase(path.begin());
        count--;

if(count == 0)
        {
            vector<int> tmp;
            vector<TreeNode *>::iterator it = path.begin();
            for(; it != path.end(); ++it)
            {
                tmp.push_back((*it)->val);
            }
            if(tmp.size() > 0)
            {
                matrix.push_back(tmp);
            }
            count = path.size();
        }
    }
    return matrix;
}

int main()
{

vector<TreeNode *> vRoot;
vector<vector<int> > ans;

vRoot = generateTrees(3);

for (int n = 0; n < vRoot.size(); ++n)
    {
        ans.clear();
        ans = levelOrder(vRoot[n]);
        cout << "----------------------" << endl;
        for (int i = 0; i < ans.size(); ++i)
        {
            for (int j = 0; j < ans[i].size(); ++j)
            {
                cout << ans[i][j] << " ";
            }
            cout << endl;
        }
    }

for (int i = 0; i < vRoot.size(); ++i)
    {
        DestroyTree(vRoot[i]);
    }
    return 0;
}

结果输出：

----------------------

1 3

----------------------

BinaryTree.h:

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#ifndef _BINARY_TREE_H_
#define _BINARY_TREE_H_

struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

TreeNode *CreateBinaryTreeNode(int value);
void ConnectTreeNodes(TreeNode *pParent,
TreeNode *pLeft, TreeNode *pRight);
void PrintTreeNode(TreeNode *pNode);
void PrintTree(TreeNode *pRoot);
void DestroyTree(TreeNode *pRoot);

#endif /*_BINARY_TREE_H_*/

BinaryTree.cpp:

#include <iostream>
#include <cstdio>
#include "BinaryTree.h"

using namespace std;

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/

//创建结点
TreeNode *CreateBinaryTreeNode(int value)
{
TreeNode *pNode = new TreeNode(value);

return pNode;
}

//连接结点
void ConnectTreeNodes(TreeNode *pParent, TreeNode *pLeft, TreeNode *pRight)
{
    if(pParent != NULL)
    {
        pParent->left = pLeft;
        pParent->right = pRight;
    }
}

//打印节点内容以及左右子结点内容
void PrintTreeNode(TreeNode *pNode)
{
    if(pNode != NULL)
    {
        printf("value of this node is: %d\n", pNode->val);

if(pNode->left != NULL)
            printf("value of its left child is: %d.\n", pNode->left->val);
        else
            printf("left child is null.\n");

if(pNode->right != NULL)
            printf("value of its right child is: %d.\n", pNode->right->val);
        else
            printf("right child is null.\n");
    }
    else
    {
        printf("this node is null.\n");
    }

printf("\n");
}

//前序遍历递归方法打印结点内容
void PrintTree(TreeNode *pRoot)
{
PrintTreeNode(pRoot);

if(pRoot != NULL)
    {
        if(pRoot->left != NULL)
            PrintTree(pRoot->left);

if(pRoot->right != NULL)
PrintTree(pRoot->right);
}
}

void DestroyTree(TreeNode *pRoot)
{
    if(pRoot != NULL)
    {
        TreeNode *pLeft = pRoot->left;
        TreeNode *pRight = pRoot->right;

delete pRoot;
pRoot = NULL;

DestroyTree(pLeft);
DestroyTree(pRight);
}
}

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【二叉查找树】02不同的二叉查找树个数II【Unique Binary Search Trees II】

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