【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; /** //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) vector<vector<int> > levelOrder(TreeNode *root) vector<vector<int> > matrix; vector<TreeNode *> path; int count = 1; if(count == 0) int main() vector<TreeNode *> vRoot; vRoot = generateTrees(3); for (int n = 0; n < vRoot.size(); ++n) for (int i = 0; i < vRoot.size(); ++i) |
|
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#ifndef _BINARY_TREE_H_
#define _BINARY_TREE_H_ struct TreeNode TreeNode *CreateBinaryTreeNode(int value); #endif /*_BINARY_TREE_H_*/ |
|
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#include <iostream>
#include <cstdio> #include "BinaryTree.h" using namespace std; /** //创建结点 return pNode; //连接结点 //打印节点内容以及左右子结点内容 if(pNode->left != NULL) if(pNode->right != NULL) printf("\n"); //前序遍历递归方法打印结点内容 if(pRoot != NULL) if(pRoot->right != NULL) void DestroyTree(TreeNode *pRoot) delete pRoot; DestroyTree(pLeft); |
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