Java

  • /**
    
     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].
    
     */
    public class Maximum_Binary_Tree {
    
        /**
         * Definition for a binary tree node.
         * public class TreeNode {
         *     int val;
         *     TreeNode left;
         *     TreeNode right;
         *     TreeNode(int x) { val = x; }
         * }
         */
        class Solution {
            public TreeNode constructMaximumBinaryTree(int[] nums) {
    
                if (nums == null || nums.length == 0) {
                    return null;
                }
    
                return construct(nums, 0, nums.length - 1);
            }
    
            private TreeNode construct(int[] nums, int start, int end) {
                if (start > end || start < 0 || end >= nums.length) {
                    return null;
                }
    
                // find max index
                int maxIndex = start;
                for (int i = start + 1; i <= end; i++) {
                    if (nums[i] > nums[maxIndex]) {
                        maxIndex = i;
                    }
                }
    
                TreeNode parent = new TreeNode(nums[maxIndex]);
                parent.left = construct(nums, start, maxIndex - 1);
                parent.right = construct(nums, maxIndex + 1, end);
    
                return parent;
            }
        }
    }
    
  • // OJ: https://leetcode.com/problems/maximum-binary-tree/
    // Time: O(N^2)
    // Space: O(N)
    class Solution {
    private:
        TreeNode *build(vector<int> &nums, int start, int end) {
            if (start >= end) return NULL;
            if (start + 1 == end) return new TreeNode(nums[start]);
            int best = start;
            for (int i = start; i < end; ++i) {
                if (nums[i] > nums[best]) best = i;
            }
            auto node = new TreeNode(nums[best]);
            node->left = build(nums, start, best);
            node->right = build(nums, best + 1, end);
            return node;
        }
    public:
        TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
            return build(nums, 0, nums.size());
        }
    };
    
  • # Definition for a binary tree node.
    # class TreeNode(object):
    #     def __init__(self, x):
    #         self.val = x
    #         self.left = None
    #         self.right = None
    
    class Solution(object):
      # recursion
      def _constructMaximumBinaryTree(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        if nums:
          pos = nums.index(max(nums))
          root = TreeNode(nums[pos])
          root.left = self.constructMaximumBinaryTree(nums[:pos])
          root.right = self.constructMaximumBinaryTree(nums[pos + 1:])
          return root
    
      # decreasing stack
      def constructMaximumBinaryTree(self, nums):
        stack = []
        for num in nums:
          root = TreeNode(num)
          while stack and stack[-1].val < num:
            root.left = stack.pop()
          if stack:
            stack[-1].right = root
          stack.append(root)
        return stack and stack[0]
    
    

Java

  • /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public TreeNode constructMaximumBinaryTree(int[] nums) {
            Stack<TreeNode> stack = new Stack<TreeNode>();
            int length = nums.length;
            for (int i = 0; i < length; i++) {
                int num = nums[i];
                TreeNode node = new TreeNode(num);
                TreeNode temp = null;
                while (!stack.isEmpty() && stack.peek().val < num)
                    temp = stack.pop();
                if (!stack.isEmpty())
                    stack.peek().right = node;
                stack.push(node);
                node.left = temp;
            }
            TreeNode root = null;
            while (!stack.isEmpty())
                root = stack.pop();
            return root;
        }
    }
    
  • // OJ: https://leetcode.com/problems/maximum-binary-tree/
    // Time: O(N^2)
    // Space: O(N)
    class Solution {
    private:
        TreeNode *build(vector<int> &nums, int start, int end) {
            if (start >= end) return NULL;
            if (start + 1 == end) return new TreeNode(nums[start]);
            int best = start;
            for (int i = start; i < end; ++i) {
                if (nums[i] > nums[best]) best = i;
            }
            auto node = new TreeNode(nums[best]);
            node->left = build(nums, start, best);
            node->right = build(nums, best + 1, end);
            return node;
        }
    public:
        TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
            return build(nums, 0, nums.size());
        }
    };
    
  • # Definition for a binary tree node.
    # class TreeNode(object):
    #     def __init__(self, x):
    #         self.val = x
    #         self.left = None
    #         self.right = None
    
    class Solution(object):
      # recursion
      def _constructMaximumBinaryTree(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        if nums:
          pos = nums.index(max(nums))
          root = TreeNode(nums[pos])
          root.left = self.constructMaximumBinaryTree(nums[:pos])
          root.right = self.constructMaximumBinaryTree(nums[pos + 1:])
          return root
    
      # decreasing stack
      def constructMaximumBinaryTree(self, nums):
        stack = []
        for num in nums:
          root = TreeNode(num)
          while stack and stack[-1].val < num:
            root.left = stack.pop()
          if stack:
            stack[-1].right = root
          stack.append(root)
        return stack and stack[0]
    
    

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