894. All Possible Full Binary Trees

梦里梦外; 2024-04-06 09:37 57阅读 0赞

Given an integer `n`, return *a list of all possible **full binary trees** with* `n` *nodes*. Each node of each tree in the answer must have `Node.val == 0`.

Each element of the answer is the root node of one possible tree. You may return the final list of trees in **any order**.

A **full binary tree** is a binary tree where each node has exactly `0` or `2` children.

**Example 1:**

![ff3f0a0345d98c1f470bb26b1014a9ea.png][]

Input: n = 7
    Output: [[0,0,0,null,null,0,0,null,null,0,0],[0,0,0,null,null,0,0,0,0],[0,0,0,0,0,0,0],[0,0,0,0,0,null,null,null,null,0,0],[0,0,0,0,0,null,null,0,0]]

**Example 2:**

Input: n = 3
    Output: [[0,0,0]]

**Constraints:**

*  `1 <= n <= 20`

题目：给定正整数n。求包含n个节点的所有完全二叉树的结构。每个节点值为0. 完全二叉树是一个二叉树，其所有节点包含0个或2个子节点。

思路：与[96. Unique Binary Search Trees][]类似，采用动态规划方法，将n-1个节点分为左右子节点。采用递归方法分别求出从1到n-2个节点的完全二叉树结构，将其置于根节点的左右子树位置。为减少计算，使用hashmap来保存子树结构。小编使用一个vector<vector<TreeNode\*>> subTree结构来代替hashmap，subTree\[i\]即为包含i个节点的所有完全二叉树结构。为保证树结构为完全二叉树，i为奇数。代码：

/**
     * 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 {
    private:
        vector<vector<TreeNode*>> subTree;
    public:
        vector<TreeNode*> allPossibleFBT(int n) {
            if(subTree.size() > n && subTree[n].size() > 0)
                return subTree[n];
            while(subTree.size() <= n) subTree.push_back({});
            if(n == 1){
                subTree[1].push_back(new TreeNode(0));
            } else {            
                for(int i = 1; i < n-1; i+=2){
                    vector<TreeNode*> left = allPossibleFBT(i);
                    vector<TreeNode*> right = allPossibleFBT(n-i-1);
                    for(auto l : left){
                        for(auto r : right){
                            TreeNode* root = new TreeNode(0);
                            root->left = l;
                            root->right = r;
                            subTree[n].push_back(root);
                        }
                    }
                }
            }
            return subTree[n];
        }
    };

[ff3f0a0345d98c1f470bb26b1014a9ea.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/06/e95883fe233b406e8b278d5d96084c9a.png
[96. Unique Binary Search Trees]: https://blog.csdn.net/qing2019/article/details/126645065?spm=1001.2014.3001.5501