AlgoDaily 10: Binary Tree In-order Traversal

“Can you write a function to traverse a binary tree in-order, and print out the value of each node as it passes?”

“The example would output [4, 6, 5] since there is no left child for 4, and 6 is visited in-order before 5.

The definition of a tree node is as follows:”

function Node(val) {
  this.val = val;
  this.left = null;
  this.right = null;
}

Another example tree:

Should result in [17, 10, 3, 5, 8].

Iterating with a stack seems right for traversing a tree in order.

From a given node, we need to get to its left-most descendent node first, pushing each node we visit on the way on to the stack to look at afterwards.

function inorderTraversal(node) {
	const result = [];
	const stack = [];
	while (true) {
		// Keep descending left until there is no left child.
		if (node) {
			stack.push(node);
			node = node.left;
			continue;
		}

		// Take the most recent node back off the stack, take its value, and
		// continue from its right child.
		if (stack.length > 0) {
			node = stack.pop();
			result.push(node.val);
			node = node.right;
			continue;
		}

		// We've visited all nodes; finish.
		break;
	}
	return result;
}

This should be O(n).

“Follow up: you’ll likely get the recursive solution first, could you do it iteratively?”

I actually did the iterative approach first. The recursive approach could look like this:

function inorderTraversal(node, result = []) {
	if (!node) {
		return result;
	}
	if (node.left) {
		inorderTraversal(node.left, result);
	}
	result.push(node.val);
	if (node.right) {
		inorderTraversal(node.right, result);
	}
	return result;
}

This is a lot more compact than the iterative version, but could exhaust the execution stack for a large tree.

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