[LeetCode] (medium) 654. Maximum Binary Tree

https://leetcode.com/problems/maximum-binary-tree/

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].

---

递归解：

/**
 * 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 construct(nums, 0, nums.size()-1);
    }
    
    TreeNode* construct(vector<int> &nums, int lo, int hi){
        if(lo > hi){
            return NULL;
        }else if(lo == hi){
            return new TreeNode(nums[lo]);
        }else{
            int M = nums[lo];
            int idx = lo;
            for(int i = lo+1; i <= hi; ++i){
                if(nums[i] > M){
                    M = nums[i];
                    idx = i;
                }
            }
            TreeNode *result = new TreeNode(M);
            result->left = construct(nums, lo, idx-1);
            result->right = construct(nums, idx+1, hi);
            return result;
        }
    }
};

 

迭代解：

保证树的中序遍历的同时确保任意结点大于他的所有子节点

class Solution {
public:
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        if(nums.empty()) return NULL;
        TreeNode* head = new TreeNode(nums[0]);
        for(int i = 1; i < nums.size(); ++i){
            if(nums[i] > head->val){
                TreeNode* nhead = new TreeNode(nums[i]);
                nhead->left = head;
                head = nhead;
            }else{
                TreeNode* cur = head;
                TreeNode* nNode = new TreeNode(nums[i]);
                while(true){
                    if(cur->right == NULL){
                        cur->right = nNode;
                        break;
                    }else if(cur->right->val < nNode->val){
                        nNode->left = cur->right;
                        cur->right = nNode;
                        break;
                    }
                    cur = cur->right;
                }
            }
        }
        
        return head;
    }
};