Maximum Binary Tree

LeetCode Q 654 - Maximum Binary Tree

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, null, 2, 0, null, null, null, null, 1]

Note: The size of the given array will be in the range [1,1000].

Solution:

Solution 1: Recursion

Code:

public TreeNode constructMaximumBinaryTree(int[] nums) {
  return dfs(nums, 0, nums.length - 1);
}

private TreeNode dfs(int[] nums, int left, int right) {
  if (left > right) return null;
  if (left == right) return new TreeNode(nums[left]);

  int maxIndex = left;
  for (int i = left; i <= right; i++) {
    if (nums[i] > nums[maxIndex]) maxIndex = i;
  }

  TreeNode n = new TreeNode(nums[maxIndex]);
  n.left = dfs(nums, left, maxIndex - 1);
  n.left = dfs(nums, maxIndex + 1, right);

  return n;
}

Solution 2: Iteration (Monotonous Stack)

Code:

public TreeNode constructMaximumBinaryTree(int[] nums) {
  Deque<TreeNode> stack = new ArrayDeque<>();
  for (int num: nums) {
    TreeNode curr = new TreeNode(num);
    while (!stack.isEmpty() && stack.peek().val < num) {
      curr.left = stack.pop();
    }
    if (!stack.isEmpty()) stack.peek().right = curr;
    stack.push(curr);
  }

  return stack.removeLast();
}