[Leetcode] 654. Maximum Binary Tree 解题报告-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/79166718

本文介绍了一种构建最大二叉树的方法，通过寻找数组中的最大元素作为根节点，并递归构造左右子树来完成整个树的构建过程。

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题目：

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].

思路：

首先找到数组中的最大元素，然后建立根节点，最后递归生成其左右子树即可。

代码：

/**
 * 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) {
        if (nums.size() == 0) {
            return NULL;
        }
        return constructMaximumBinaryTree(nums, 0, nums.size() - 1);
    }
private:
    TreeNode* constructMaximumBinaryTree(vector<int> &nums, int left, int right) {
        if (left > right) {
            return NULL;
        }
        else {
            int max_index = getMaxIndex(nums, left, right);
            TreeNode *root = new TreeNode(nums[max_index]);
            root->left = constructMaximumBinaryTree(nums, left, max_index - 1);
            root->right = constructMaximumBinaryTree(nums, max_index + 1, right);
            return root;
        }
    }
    int getMaxIndex(vector<int> &nums, int left, int right) {
        int max_index = left, max_value = nums[left];
        for (int i = left + 1; i <= right; ++i) {
            if (nums[i] > max_value) {
                max_value = nums[i];
                max_index = i;
            }
        }
        return max_index;
    }
};