95. Unique Binary Search Trees II

Description

Given an integer n, return all the structurally unique BST's (binary search trees), which has exactly n nodes of unique values from 1 to n. Return the answer in any order.

Example 1:

Input: n = 3
Output: [[1,null,2,null,3],[1,null,3,2],[2,1,3],[3,1,null,null,2],[3,2,null,1]]

Example 2:

Input: n = 1
Output: [[1]]

Constraints:

1 <= n <= 8

Solutions

Solution 1: DFS (Depth-First Search)

We design a function $dfs(i, j)$ that returns all feasible binary search trees composed of $[i, j]$, so the answer is $dfs(1, n)$.

The execution steps of the function $dfs(i, j)$ are as follows:

If $i > j$, it means that there are no numbers to form a binary search tree at this time, so return a list consisting of a null node.
If $i \leq j$, we enumerate the numbers $v$ in $[i, j]$ as the root node. The left subtree of the root node $v$ is composed of $[i, v - 1]$, and the right subtree is composed of $[v + 1, j]$. Finally, we take the Cartesian product of all combinations of the left and right subtrees, i.e., $left \times right$, add the root node $v$, and get all binary search trees with $v$ as the root node.

The time complexity is $O(n \times G(n))$, and the space complexity is $O(n \times G(n))$. Where $G(n)$ is the Catalan number.

Python3JavaC++GoTypeScriptRust

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def generateTrees(self, n: int) -> List[Optional[TreeNode]]:
        def dfs(i: int, j: int) -> List[Optional[TreeNode]]:
            if i > j:
                return [None]
            ans = []
            for v in range(i, j + 1):
                left = dfs(i, v - 1)
                right = dfs(v + 1, j)
                for l in left:
                    for r in right:
                        ans.append(TreeNode(v, l, r))
            return ans

        return dfs(1, n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<TreeNode> generateTrees(int n) {
        return dfs(1, n);
    }

    private List<TreeNode> dfs(int i, int j) {
        List<TreeNode> ans = new ArrayList<>();
        if (i > j) {
            ans.add(null);
            return ans;
        }
        for (int v = i; v <= j; ++v) {
            var left = dfs(i, v - 1);
            var right = dfs(v + 1, j);
            for (var l : left) {
                for (var r : right) {
                    ans.add(new TreeNode(v, l, r));
                }
            }
        }
        return ans;
    }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<TreeNode*> generateTrees(int n) {
        function<vector<TreeNode*>(int, int)> dfs = [&](int i, int j) {
            if (i > j) {
                return vector<TreeNode*>{nullptr};
            }
            vector<TreeNode*> ans;
            for (int v = i; v <= j; ++v) {
                auto left = dfs(i, v - 1);
                auto right = dfs(v + 1, j);
                for (auto l : left) {
                    for (auto r : right) {
                        ans.push_back(new TreeNode(v, l, r));
                    }
                }
            }
            return ans;
        };
        return dfs(1, n);
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func generateTrees(n int) []*TreeNode {
    var dfs func(int, int) []*TreeNode
    dfs = func(i, j int) []*TreeNode {
        if i > j {
            return []*TreeNode{nil}
        }
        ans := []*TreeNode{}
        for v := i; v <= j; v++ {
            left := dfs(i, v-1)
            right := dfs(v+1, j)
            for _, l := range left {
                for _, r := range right {
                    ans = append(ans, &TreeNode{v, l, r})
                }
            }
        }
        return ans
    }
    return dfs(1, n)
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function generateTrees(n: number): Array<TreeNode | null> {
    const dfs = (i: number, j: number): Array<TreeNode | null> => {
        if (i > j) {
            return [null];
        }
        const ans: Array<TreeNode | null> = [];
        for (let v = i; v <= j; ++v) {
            const left = dfs(i, v - 1);
            const right = dfs(v + 1, j);
            for (const l of left) {
                for (const r of right) {
                    ans.push(new TreeNode(v, l, r));
                }
            }
        }
        return ans;
    };
    return dfs(1, n);
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    pub fn generate_trees(n: i32) -> Vec<Option<Rc<RefCell<TreeNode>>>> {
        Self::dfs(1, n)
    }

    fn dfs(i: i32, j: i32) -> Vec<Option<Rc<RefCell<TreeNode>>>> {
        let mut ans = Vec::new();
        if i > j {
            ans.push(None);
            return ans;
        }
        for v in i..=j {
            let left = Self::dfs(i, v - 1);
            let right = Self::dfs(v + 1, j);
            for l in &left {
                for r in &right {
                    ans.push(Some(Rc::new(RefCell::new(TreeNode {
                        val: v,
                        left: l.clone(),
                        right: r.clone(),
                    }))));
                }
            }
        }
        ans
    }
}

95. Unique Binary Search Trees II

Description

Solutions

Solution 1: DFS (Depth-First Search)

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