Given an integer array with no duplicates. A maximum tree building on this array is defined as follow:

  1. The root is the maximum number in the array.
  2. The left subtree is the maximum tree constructed from left part subarray divided by the maximum number.
  3. The right subtree is the maximum tree constructed from right part subarray divided by the maximum number.

Construct the maximum tree by the given array and output the root node of this tree.

Example 1:

Input: [3,2,1,6,0,5]
Output: return the tree root node representing the following tree: 6
/ \
3 5
\ /
2 0
\
1

Note:

  1. The size of the given array will be in the range [1,1000].

给一个数组,以数组中的最大值为根结点创建一个最大二叉树,分隔出的左右部分再分别创建最大二叉树。

解法:递归

Java:

public class Solution {
public TreeNode constructMaximumBinaryTree(int[] nums) {
if (nums == null) return null;
return build(nums, 0, nums.length - 1);
} private TreeNode build(int[] nums, int start, int end) {
if (start > end) return null; int idxMax = start;
for (int i = start + 1; i <= end; i++) {
if (nums[i] > nums[idxMax]) {
idxMax = i;
}
} TreeNode root = new TreeNode(nums[idxMax]); root.left = build(nums, start, idxMax - 1);
root.right = build(nums, idxMax + 1, end); return root;
}
}

Java:

public TreeNode constructMaximumBinaryTree(int[] nums) {
return construct(nums, 0, nums.length);
} TreeNode construct(int[] nums, int l, int r) {
if (l >= r) return null;
int maxi = l;
for (int i = l + 1; i < r; i++) if (nums[i] > nums[maxi]) maxi = i;
TreeNode root = new TreeNode(nums[maxi]);
root.left = construct(nums, l, maxi);
root.right = construct(nums, maxi + 1, r);
return root;
}  

Python:

def constructMaximumBinaryTree(self, nums):
if not nums:
return None
root, maxi = TreeNode(max(nums)), nums.index(max(nums))
root.left = self.constructMaximumBinaryTree(nums[:maxi])
root.right = self.constructMaximumBinaryTree(nums[maxi + 1:])
return root  

Python:

# Time:  O(n)
# Space: O(n)
class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
self.right = None class Solution(object):
def constructMaximumBinaryTree(self, nums):
"""
:type nums: List[int]
:rtype: TreeNode
"""
nodeStack = []
for num in nums:
node = TreeNode(num);
while nodeStack and num > nodeStack[-1].val:
node.left = nodeStack.pop()
if nodeStack:
nodeStack[-1].right = node
nodeStack.append(node)
return nodeStack[0]

Python:

class Solution(object):
def constructMaximumBinaryTree(self, nums):
"""
:type nums: List[int]
:rtype: TreeNode
"""
if not nums:
return mx = float('-inf')
mx_index = 0
for i in range(len(nums)):
if nums[i] > mx:
mx = nums[i]
mx_index = i root = TreeNode(mx)
if mx_index > 0:
root.left = self.constructMaximumBinaryTree(nums[:mx_index])
if mx_index < len(nums) - 1:
root.right = self.constructMaximumBinaryTree(nums[mx_index+1:]) return root  

C++:

class Solution {
public:
TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
if (nums.empty()) return NULL;
int mx = INT_MIN, mx_idx = 0;
for (int i = 0; i < nums.size(); ++i) {
if (mx < nums[i]) {
mx = nums[i];
mx_idx = i;
}
}
TreeNode *node = new TreeNode(mx);
vector<int> leftArr = vector<int>(nums.begin(), nums.begin() + mx_idx);
vector<int> rightArr = vector<int>(nums.begin() + mx_idx + 1, nums.end());
node->left = constructMaximumBinaryTree(leftArr);
node->right = constructMaximumBinaryTree(rightArr);
return node;
}
};

C++:  

class Solution {
public:
TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
if (nums.empty()) return NULL;
return helper(nums, 0, nums.size() - 1);
}
TreeNode* helper(vector<int>& nums, int left, int right) {
if (left > right) return NULL;
int mid = left;
for (int i = left + 1; i <= right; ++i) {
if (nums[i] > nums[mid]) {
mid = i;
}
}
TreeNode *node = new TreeNode(nums[mid]);
node->left = helper(nums, left, mid - 1);
node->right = helper(nums, mid + 1, right);
return node;
}
};

C++:  

class Solution {
public:
TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
vector<TreeNode*> v;
for (int num : nums) {
TreeNode *cur = new TreeNode(num);
while (!v.empty() && v.back()->val < num) {
cur->left = v.back();
v.pop_back();
}
if (!v.empty()) {
v.back()->right = cur;
}
v.push_back(cur);
}
return v.front();
}
};

  

  

  

All LeetCode Questions List 题目汇总

[LeetCode] 654. Maximum Binary Tree 最大二叉树的更多相关文章

  1. LeetCode 654. Maximum Binary Tree最大二叉树 (C++)

    题目: Given an integer array with no duplicates. A maximum tree building on this array is defined as f ...

  2. LeetCode - 654. Maximum Binary Tree

    Given an integer array with no duplicates. A maximum tree building on this array is defined as follo ...

  3. 654. Maximum Binary Tree最大二叉树

    网址:https://leetcode.com/problems/maximum-binary-tree/ 参考: https://leetcode.com/problems/maximum-bina ...

  4. [LeetCode]654. Maximum Binary Tree最大堆二叉树

    每次找到数组中的最大值,然后递归的构建左右树 public TreeNode constructMaximumBinaryTree(int[] nums) { if (nums.length==0) ...

  5. 654. Maximum Binary Tree

    654. Maximum Binary Tree 题目大意: 意思就是给你一组数,先选一个最大的作为根,这个数左边的数组作为左子树,右边的数组作为右子树,重复上一步. 读完就知道是递归了. 这个题真尼 ...

  6. [Leetcode Week14]Maximum Binary Tree

    Maximum Binary Tree 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/maximum-binary-tree/description/ ...

  7. 【LeetCode】654. Maximum Binary Tree 解题报告 (Python&C++)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 递归 日期 题目地址:https://leetcode ...

  8. 【leetcode】654. Maximum Binary Tree

    题目如下: Given an integer array with no duplicates. A maximum tree building on this array is defined as ...

  9. [LeetCode] 655. Print Binary Tree 打印二叉树

    Print a binary tree in an m*n 2D string array following these rules: The row number m should be equa ...

随机推荐

  1. 大数据开发之keras代码框架应用

    总体来讲keras这个深度学习框架真的很“简易”,它体现在可参考的文档写的比较详细,不像caffe,装完以后都得靠技术博客,keras有它自己的官方文档(不过是英文的),这给初学者提供了很大的学习空间 ...

  2. async-validator 表单验证注意事项

    1. this.$refs[formName].validate()里面的内容不执行 解决问题出处:https://segmentfault.com/q/1010000009679079 问题描述:1 ...

  3. 7-html列表

    <!DOCTYPE html> <html> <head> <meta charset="utf-8"> <meta http ...

  4. Task 使用方法

    Task的使用方法 1. 调用无参数.无返回值方法 private void button1_Click(object sender, EventArgs e) { Task task = new T ...

  5. FFT版题 [51 Nod 1028] 大数乘法

    题目链接:51 Nod 传送门 数的长度为10510^5105,乘起来后最大长度为2×1052\times10^52×105 由于FFT需要把长度开到222的次幂,所以不能只开到2×1052\time ...

  6. B/S结构与C/S结构测试区别

    B/S结构与C/S结构 B/S结构是浏览器/服务器结构,应用软件的业务逻辑完全在服务器端实现,客户端只需要通过浏览器完成浏览.查询.输入等简单操作. C/S结构是客户端/浏览器结构,客户端具有一定的数 ...

  7. 使用terraform 进行gitlab 代码仓库批量迁移

      gitlab 的代码是在文件目录中,这个对于批量迁移很简单,只需要copy 文件夹(但是对于不同gitlab server 可能需要重新设置目录权限) 几个问题 大批量仓库tf resource问 ...

  8. render函数、createElement函数与vm.$slots

    1.render函数.createElement函数 Vue.component('es-header', { render: function (createElement) { return cr ...

  9. 初识 Python 作业及默写

    1.简述变量量命名规范 2.name = input(“>>>”) name变量是什么数据类型? 3.if条件语句的基本结构? 4.用print打印出下面内容: 文能提笔安天下, 武 ...

  10. GoCN每日新闻(2019-09-30)

    GoCN每日新闻(2019-09-30) 1. 使用Sqlmock测试数据库 https://medium.com/ralali-engineering/testing-database-using- ...