### Container with Most Water - Leet Code Solution

Problem Statement Given n non-negative integers a1, a2, …, an , where each…

November 19, 2020

Given a Binary tree, print out nodes in level order traversal from left to right.

```
50
/ \
80 30
/ \ \
20 40 10
# Output
# 50 80 30 20 40 10
```

Use BFS (Breadth First Search) algorithm. Since, it reaches out to nodes first that are immediate neighbours.

Idea is to take a queue, keep accumulating queue for each child.

```
void bfs(Node node) {
Queue<Node> q = new Queue();
q.add(node);
while (!q.isEmpty()) {
Node n = q.pop();
print(n);
if (n.left != null) q.add(n.left);
if (n.right != null) q.add(n.right);
}
}
```

The `Complexity`

is `O(n)`

as we are visiting each nodes only once.

You can prepare a list of nodes at each level. We will also use DFS (Depth First Search) algorithm here.

```
void levelOrder(Node node, List<List<Node>> list, int level) {
if (node == null)
return;
List<Node> levelList = list.get(level);
if (levelList == null) {
levelList = new ArrayList();
list.add(levelList);
}
levelList.add(node);
levelOrder(node.left, list, level+1);
levelOrder(node.right, list, level+1);
}
List<List<Node>> list = new ArrayList();
levelOrder(root, list, 0);
```

After this, we can print the list.

The `Complexity`

is `O(n)`

as we are visiting each nodes only once.

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