算法第七周Maximum Binary Tree[medium]

Maximum Binary Tree[medium]

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Description

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

The root is the maximum number in the array.
The left subtree is the maximum tree constructed from left part subarray divided by the maximum number.
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:
The size of the given array will be in the range [1,1000].

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Solution

这道题目是用来构建最大二叉树，其意思是我们在给定数组中选择最大的数作为根节点，然后将位于其左边的数字（数组中）作为它的左子树，右边的数字作为它的有右字数。显然我们可以通过递归的方式构建，通过left和right变量记录每次数组的大小。每次递归函数中首先寻找最大的数字，然后将数组分组继续递归。

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {        TreeNode* result = MaxTree(nums, 0, nums.size());        return result;    }    TreeNode* MaxTree(vector<int>& nums, int left, int right) {        if (right <= left) return NULL;        TreeNode* root;        int index;        int max = INT_MIN;        for (int i = left; i < right; i++) {            if (nums[i] > max) {                max = nums[i];                index = i;            }        }        root = new TreeNode(max);        root->left = MaxTree(nums, left, index);        root->right = MaxTree(nums, index+1, right);        return root;    }};