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.

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​

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)

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