Good morning! Here's our prompt for today.

Assume we're using a binary tree in writing a video game via Binary Space Partitioning. We need to identify the bottom left leaf, that is-- the leftmost value in the lowest row of the binary tree.

In this example, the bottom left leaf is 3:

1/*
2    4
3   / \
4  3   5
5*/

Assuming the standard node definition of:

1function Node(val) {
2  this.val = val;
3  this.left = this.right = null;
4}

It would be called as such:

1const root = new Node(4);
2root.left = new Node(3);
3root.right = new Node(5);
4
5function bottomLeftNodeVal(root) { return; };
6bottomLeftNodeVal(root);
7// 3

Here's another example. Let's assume that there's at least the root node available in all binary trees passed as parameters.

1/*
2      4
3     / \
4    1   3
5       / \
6      5   9
7*/
8
9const root = new Node(4);
10root.left = new Node(1);
11root.right = new Node(3);
12root.right.left = new Node(5);
13root.right.right = new Node(9);
14
15bottomLeftNodeVal(root);
16// 5

Constraints

• Nodes in the binary tree <= 100000
• The values in the nodes will be in the range -1000000000 and 1000000000
• Expected time complexity : O(n)
• Expected space complexity : O(n)

Try to solve this here or in Interactive Mode.

How do I practice this challenge?

xxxxxxxxxx

​

def bottom_left_node(root):

    # fill in

    return root

​

​

# Node definition

class Node:

    def __init__(self, val):

        self.val = val

        self.left = None

        self.right = None

​

​

# Regular binary trees

tree1 = Node(4)

tree1.left = Node(1)

tree1.right = Node(3)

​

tree2 = Node(5)

tree2.left = Node(10)

tree2.left.left = Node(17)

tree2.left.right = Node(3)

tree2.right = Node(8)

​

# Binary search trees

tree3 = Node(6)

tree3.left = Node(3)

​

tree4 = Node(5)

Here's a video of us explaining the solution.

To change the speed of the video or see it in full screen, click the icons to the right of the progress bar.

We'll now take you through what you need to know.

How do I use this guide?