一、题目

在这里插入图片描述

二、解法

思路分析：【算法与数据结构】106、LeetCode从中序与后序遍历序列构造二叉树这两道题有些类似，相关代码可以互相参考，本题明示了要用递归来做，那么递归三要素不可缺少：输入参数和返回值；单层递归逻辑；终止条件。本题当中，输入参数引用二叉树遍历数组，同时根据最大值划分的边界[Begin, End)，代码统一为左闭右开区间，区间具体如何划分设计到边界条件是否+1-1这种。返回值为root根节点。
程序如下：

class Solution {
public:// 3、输入参数TreeNode* traversal(const vector<int> &nums, int Begin, int End) {// 1、终止条件if (Begin == End) return NULL;//2、单层递归逻辑int maxIndex = Begin;for (int i = Begin; i < End; i++) { // 找最大值if (nums[i] > nums[maxIndex]) maxIndex = i;}TreeNode* root = new TreeNode(nums[maxIndex]);// 最大值左边部分，左闭右开[leftBegin, leftEnd)int leftBegin = Begin; int leftEnd = maxIndex;if (leftEnd < leftBegin) leftEnd = Begin;// 最大值右边部分，左闭右开[rightBegin, rightEnd)int rightBegin = maxIndex + 1;int rightEnd = End;if (rightBegin > rightEnd) rightBegin = End;root->left = traversal(nums, leftBegin, leftEnd);root->right = traversal(nums, rightBegin, rightEnd);// 3、返回值return root;}TreeNode* constructMaximumBinaryTree(vector<int>& nums) {  if (!nums.size()) return NULL;return traversal(nums, 0, nums.size());}
};

三、完整代码

# include <iostream>
# include <vector>
# include <queue>
using namespace std;// 树节点定义
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:// 3、输入参数TreeNode* traversal(const vector<int> &nums, int Begin, int End) {// 1、终止条件if (Begin == End) return NULL;//2、单层递归逻辑int maxIndex = Begin;for (int i = Begin; i < End; i++) { // 找最大值if (nums[i] > nums[maxIndex]) maxIndex = i;}TreeNode* root = new TreeNode(nums[maxIndex]);// 最大值左边部分，左闭右开[leftBegin, leftEnd)int leftBegin = Begin; int leftEnd = maxIndex;if (leftEnd < leftBegin) leftEnd = Begin;// 最大值右边部分，左闭右开[rightBegin, rightEnd)int rightBegin = maxIndex + 1;int rightEnd = End;if (rightBegin > rightEnd) rightBegin = End;root->left = traversal(nums, leftBegin, leftEnd);root->right = traversal(nums, rightBegin, rightEnd);// 3、返回值return root;}TreeNode* constructMaximumBinaryTree(vector<int>& nums) {  if (!nums.size()) return NULL;return traversal(nums, 0, nums.size());}
};// 层序遍历
vector<vector<int>> levelOrder(TreeNode* root) {queue<TreeNode*> que;if (root != NULL) que.push(root);vector<vector<int>> result;while (!que.empty()) {int size = que.size();  // size必须固定, que.size()是不断变化的vector<int> vec;for (int i = 0; i < size; ++i) {TreeNode* node = que.front();que.pop();vec.push_back(node->val);if (node->left) que.push(node->left);if (node->right) que.push(node->right);}result.push_back(vec);}return result;
}template<class T1, class T2>
void my_print2(T1& v, const string str) {cout << str << endl;for (class T1::iterator vit = v.begin(); vit < v.end(); ++vit) {for (class T2::iterator it = (*vit).begin(); it < (*vit).end(); ++it) {cout << *it << ' ';}cout << endl;}
}int main()
{//int arr[] = {3, 2, 1, 6, 0, 5};int arr[] = { 3, 2, 1};vector<int> nums(arr, arr + sizeof(arr) / sizeof(int));Solution s;TreeNode* root = s.constructMaximumBinaryTree(nums);vector<vector<int>> tree = levelOrder(root);my_print2<vector<vector<int>>, vector<int>>(tree, "目标树:");system("pause");return 0;
}