[leetcode] 654. Maximum Binary Tree

Question :

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

Solution :

按照规则设计的一个递归算法，先找到数组最大值，分成两部分，递归得到左子树和右子树。

返回的条件是遇到了最左或者最右或左==右（即父节点就是最左或最右）就返回空。若左+1==右则表示只有一个元素，返回该元素的节点。

/** * 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) {        return helper(nums, 0, nums.size());    }    TreeNode* helper(vector<int>& nums, int left, int right) {        if (left < 0 || right > nums.size() || left == right) return NULL;        if (left + 1 == right) {            TreeNode *ret = new TreeNode(nums[left]);            return ret;        }        int index = left;        for (int i = left + 1; i < right; i++) {            if (nums[i] > nums[index]) {                index = i;            }        }        TreeNode *ret = new TreeNode(nums[index]);        ret->left = helper(nums, left, index);        ret->right = helper(nums, index+1, right);        return ret;    }};