LeetCode 998: Maximum Binary Tree II Solution

Master LeetCode problem 998 (Maximum Binary Tree II), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

998. Maximum Binary Tree II

Medium

Tree Binary Tree

Problem Explanation

Explanation

To solve this problem, we can perform a recursive approach. We need to handle three cases:

If the root is null, we create a new node with the given value val and return it.
If the given value val is greater than the root's value, we create a new node with value val, where the left child is the current root and the right child is null.
If the given value val is less than the root's value, we recursively call the function on the right child of the current root until we find a suitable position to insert the new node.

The time complexity of this solution is O(n) in the worst case, where n is the number of nodes in the binary tree. The space complexity is also O(n) due to the recursive calls.

Solution Code

class Solution {
    public TreeNode insertIntoMaxTree(TreeNode root, int val) {
        if (root == null) {
            return new TreeNode(val);
        }
        
        if (val > root.val) {
            TreeNode newNode = new TreeNode(val);
            newNode.left = root;
            return newNode;
        } else {
            root.right = insertIntoMaxTree(root.right, val);
            return root;
        }
    }
}

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