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

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].

给一个数组，以数组中的最大值为根结点创建一个最大二叉树，分隔出的左右部分再分别创建最大二叉树。

解法：递归

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;
    }
}

Java:

public TreeNode constructMaximumBinaryTree(int[] nums) {
        return construct(nums, 0, nums.length);
}

TreeNode construct(int[] nums, int l, int r) {
        if (l >= r) return null;
        int maxi = l;
        for (int i = l + 1; i < r; i++) if (nums[i] > nums[maxi]) maxi = i;
        TreeNode root = new TreeNode(nums[maxi]);
        root.left = construct(nums, l, maxi);
        root.right = construct(nums, maxi + 1, r);
        return root;
}　　

Python:

def constructMaximumBinaryTree(self, nums):
        if not nums:
            return None
        root, maxi = TreeNode(max(nums)), nums.index(max(nums))
        root.left = self.constructMaximumBinaryTree(nums[:maxi])
        root.right = self.constructMaximumBinaryTree(nums[maxi + 1:])
        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
        """
        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]

Python:

class Solution(object):
    def constructMaximumBinaryTree(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        if not nums:
            return

        mx = float('-inf')
        mx_index = 0
        for i in range(len(nums)):
            if nums[i] > mx:
                mx = nums[i]
                mx_index = i
       
        root = TreeNode(mx)
        if mx_index > 0:
            root.left = self.constructMaximumBinaryTree(nums[:mx_index])
        if mx_index < len(nums) - 1:
            root.right = self.constructMaximumBinaryTree(nums[mx_index+1:])
        
        return root　　

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();
    }
};

　　

　　

　　

 

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