标签：define empty 拒绝 i++ nod desc tps get ble
原创文章，拒绝转载
题目来源：https://leetcode.com/problems/maximum-binary-tree/description/
Given an integer array with no duplicates. A maximum tree building on this array is defined as follow:
Construct the maximum tree by the given array and output the root node of this tree.
Input: [3,2,1,6,0,5]
Output: return the tree root node representing the following tree:
6
/ 3 5
\ /
2 0
1Note: The size of the given array will be in the range [1,1000].
class Solution {
public:
TreeNode* getSubTree(vector<int>& nums, int start, int end) {
TreeNode* resultNode;
if (start == end) {
resultNode = new TreeNode(nums[start]);
return resultNode;
}
int maxIdx = start;
int i;
for (i = start; i <= end; i++) {
if (nums[i] > nums[maxIdx])
maxIdx = i;
}
resultNode = new TreeNode(nums[maxIdx]);
if (maxIdx > start) {
resultNode -> left = getSubTree(nums, start, maxIdx - 1);
}
if (maxIdx < end) {
resultNode -> right = getSubTree(nums, maxIdx + 1, end);
}
return resultNode;
}
TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
if (nums.empty())
return NULL;
return getSubTree(nums, 0, nums.size() - 1);
}
};这道题的题意是，对给定的一个数组，构造一棵所谓的“最大二叉树”。很容易想到的就是使用递归的思想，每次都对数组的一段进行处理，找出数组段中最大的元素，将该元素所谓当前树的树根，对元素左右两边两个数组段分别构造“最大二叉树”，分别作为树根的左子树和右子树。
[Leetcode Week14]Maximum Binary Tree
标签：define empty 拒绝 i++ nod desc tps get ble
原文地址：http://www.cnblogs.com/yanhewu/p/7994417.html
