https://leetcode.com/problems/maximum-binary-tree-ii/

We are given the `root` node of a maximum tree: a tree where every node has a value greater than any other value in its subtree.

Just as in the previous problem, the given tree was constructed from an list `A` (`root = Construct(A)`) recursively with the following `Construct(A)` routine:

• If `A` is empty, return `null`.
• Otherwise, let `A[i]` be the largest element of `A`.  Create a `root` node with value `A[i]`.
• The left child of `root` will be `Construct([A[0], A[1], ..., A[i-1]])`
• The right child of `root` will be `Construct([A[i+1], A[i+2], ..., A[A.length - 1]])`
• Return `root`.

Note that we were not given A directly, only a root node `root = Construct(A)`.

Suppose `B` is a copy of `A` with the value `val` appended to it.  It is guaranteed that `B` has unique values.

Return `Construct(B)`.

Example 1:

```Input: root = [4,1,3,null,null,2], val = 5
Output: [5,4,null,1,3,null,null,2]
Explanation: A = [1,4,2,3], B = [1,4,2,3,5]
```

Example 2:

```Input: root = [5,2,4,null,1], val = 3
Output: [5,2,4,null,1,null,3]
Explanation: A = [2,1,5,4], B = [2,1,5,4,3]
```

Example 3:

```Input: root = [5,2,3,null,1], val = 4
Output: [5,2,4,null,1,3]
Explanation: A = [2,1,5,3], B = [2,1,5,3,4]
```

Constraints:

• `1 <= B.length <= 100`

## Solution:

### Recursive:

O(N) time, O(N) space

```public TreeNode insertIntoMaxTree(TreeNode root, int val) {
if (root != null && root.val > val) {
root.right = insertIntoMaxTree(root.right, val);
return root;
}
TreeNode node = new TreeNode(val);
node.left = root;
return node;
}```
```TreeNode* insertIntoMaxTree(TreeNode* root, int val) {
if (root && root->val > val) {
root->right = insertIntoMaxTree(root->right, val);
return root;
}
TreeNode* node = new TreeNode(val);
node->left = root;
return node;
}```
```def insertIntoMaxTree(self, root: TreeNode, val: int) -> TreeNode:
if not root: return TreeNode(val)
if val > root.val :
node = TreeNode(val)
node.left = root
return node
else:
root.right = self.insertIntoMaxTree(root.right, val)
return root```

### Iterative:

`O(N)` time`, O(1)` space

```public TreeNode insertIntoMaxTree(TreeNode root, int val) {
TreeNode node = new TreeNode(val), curr = root;
if (root.val < val) {
node.left = root;
return node;
}
while (curr.right != null && curr.right.val > val) {
curr = curr.right;
}
node.left = curr.right;
curr.right = node;
return root;
}```
```TreeNode* insertIntoMaxTree(TreeNode* root, int val) {

TreeNode* tmp = root;
TreeNode* tmp2 = nullptr;
TreeNode* new_node = new TreeNode(val);
bool insert_new_node = false;

while(tmp != nullptr) {
if(tmp->val < val) {

new_node->left = tmp;
if(tmp == root)
return new_node;

tmp2->right = new_node;
insert_new_node = true;
break;
}
tmp2 = tmp;
tmp = tmp->right;
}

if(!insert_new_node) {
tmp2->right = new_node;
}

return root;
}```
```def insertIntoMaxTree(self, root, val):
pre,cur = None, root
while cur and cur.val > val:
pre, cur = cur, cur.right
node = TreeNode(val)
node.left = cur
if pre: pre.right = node
return root if root.val > val else node```
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