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index.js
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index.js
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/*
* @lc app=leetcode id=95 lang=javascript
*
* [95] Unique Binary Search Trees II
*
* https://leetcode.com/problems/unique-binary-search-trees-ii/description/
*
* algorithms
* Medium (34.56%)
* Total Accepted: 127.7K
* Total Submissions: 367.9K
* Testcase Example: '3'
*
* Given an integer n, generate all structurally unique BST's (binary search
* trees) that store values 1 ... n.
*
* Example:
*
*
* Input: 3
* Output:
* [
* [1,null,3,2],
* [3,2,null,1],
* [3,1,null,null,2],
* [2,1,3],
* [1,null,2,null,3]
* ]
* Explanation:
* The above output corresponds to the 5 unique BST's shown below:
*
* 1 3 3 2 1
* \ / / / \ \
* 3 2 1 1 3 2
* / / \ \
* 2 1 2 3
*
*
*/
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
const { TreeNode } = require('../utils');
/**
* 思路:
*
* 1.以每一个节点做根节点向下伸展
* 2.伸展时,如果左节点为空,右节点补位空,则左节点需要补充一个null
*
* @param {number} n
* @return {TreeNode[]}
*/
var generateTrees = function(n) {
if (n === 0) {
return [];
}
return getBSTs(1, n);
function getBSTs(min, max) {
const res = [];
if (min > max) {
return [null];
}
if (min === max) {
return [new TreeNode(min)];
}
for (let i = min; i <= max; i++) {
const leftRoots = getBSTs(min, i - 1);
const rightRoots = getBSTs(i + 1, max);
for (let j = 0; j < leftRoots.length; j++) {
for (let k = 0; k < rightRoots.length; k++) {
const root = new TreeNode(i);
root.left = leftRoots[j];
root.right = rightRoots[k];
res.push(root);
}
}
}
return res;
}
console.log(nums)
};
console.log(generateTrees(1));
module.exports = {
id:'95',
title:'Unique Binary Search Trees II',
url:'https://leetcode.com/problems/unique-binary-search-trees-ii/description/',
difficulty:'Medium',
}