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].

給一個數組，以數組中的最大值為根結點創建一個最大二叉樹，分隔出的左右部分再分別創建最大二叉樹。

解法：遞歸

Java:

public class Solution {
    public TreeNode constructMaximumBinaryTree(int[] nums) {
        if (nums == null) return null;
        return build(nums, 0, nums.length - 1);
    }
    
    private TreeNode build(int[] nums, int start, int end) {
        if (start > end) return null;
        
        int idxMax = start;
        for (int i = start + 1; i <= end; i++) {
            if (nums[i] > nums[idxMax]) {
                idxMax = i;
            }
        }
        
        TreeNode root = new TreeNode(nums[idxMax]);
        
        root.left = build(nums, start, idxMax - 1);
        root.right = build(nums, idxMax + 1, end);
        
        return root;
    }
}

Python:

# Time:  O(n)
# Space: O(n)
class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None


class Solution(object):
    def constructMaximumBinaryTree(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        # https://github.com/kamyu104/LintCode/blob/master/C++/max-tree.cpp
        nodeStack = []
        for num in nums:
            node = TreeNode(num);
            while nodeStack and num > nodeStack[-1].val:
                node.left = nodeStack.pop()
            if nodeStack:
                nodeStack[-1].right = node
            nodeStack.append(node)
        return nodeStack[0]

C++:

class Solution {
public:
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        if (nums.empty()) return NULL;
        int mx = INT_MIN, mx_idx = 0;
        for (int i = 0; i < nums.size(); ++i) {
            if (mx < nums[i]) {
                mx = nums[i];
                mx_idx = i;
            }
        }
        TreeNode *node = new TreeNode(mx);
        vector<int> leftArr = vector<int>(nums.begin(), nums.begin() + mx_idx);
        vector<int> rightArr = vector<int>(nums.begin() + mx_idx + 1, nums.end());
        node->left = constructMaximumBinaryTree(leftArr);
        node->right = constructMaximumBinaryTree(rightArr);
        return node;
    }
};

C++:　　

class Solution {
public:
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        if (nums.empty()) return NULL;
        return helper(nums, 0, nums.size() - 1);
    }
    TreeNode* helper(vector<int>& nums, int left, int right) {
        if (left > right) return NULL;
        int mid = left;
        for (int i = left + 1; i <= right; ++i) {
            if (nums[i] > nums[mid]) {
                mid = i;
            }
        }
        TreeNode *node = new TreeNode(nums[mid]);
        node->left = helper(nums, left, mid - 1);
        node->right = helper(nums, mid + 1, right);
        return node;
    }
};

C++:　　

class Solution {
public:
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        vector<TreeNode*> v;
        for (int num : nums) {
            TreeNode *cur = new TreeNode(num);
            while (!v.empty() && v.back()->val < num) {
                cur->left = v.back();
                v.pop_back();
            }
            if (!v.empty()) {
                v.back()->right = cur;
            }
            v.push_back(cur);
        }
        return v.front();
    }
};

[LeetCode] 654. Maximum Binary Tree 最大二叉樹