Discussion on Day 23: BST Level-Order Traversal Challenge

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430 Discussions

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1 week ago+ 0 comments

python 3 solution

def levelOrder(self,root):
    #Write your code here
    queue = [root]

    while queue:
        node = queue.pop(0)  # Pop from the front

        print(f"{node.data}", end=" ")

        for new in [node.left, node.right]:
            if new:
                queue.append(new)

4 weeks ago+ 0 comments

Since we can't use Python's collections.deque and the O(1) popleft function, it's actually more effecient to just maintain a pointer to your current location in the list (this avoids the O(N) cost of each pop(0) call).

Python 3:

    def levelOrder(self,root):
        q = []
        if root is not None:
            q.append(root)
        i = 0
        while i < len(q):
            n = q[i]
            if n.left is not None:
                q.append(n.left)
            if n.right is not None:
                q.append(n.right)
            i += 1
        print(" ".join(str(x.data) for x in q))

4 months ago+ 0 comments

JavaSCript

this.levelOrder = function(root) {
    if (!root) {
        return;
    }

    const queue = [];
    queue.push(root);

    while (queue.length > 0) {
        const node = queue.shift();
        process.stdout.write(node.data + " ");

        if (node.left) {
            queue.push(node.left);
        }
        if (node.right) {
            queue.push(node.right);
        }
    }
};

4 months ago+ 0 comments

in JS

// Start of function levelOrder
    this.levelOrder = function(root) {
    if (!root) return;

        const queue = [];
        queue.push(root);

        while (queue.length > 0) { 
            const levelSize = queue.length; 

        for (let i = 0; i < levelSize; i++) { 
        const node = queue.shift(); 
        process.stdout.write(node.data + ' '); 

        if (node.left) queue.push(node.left); 
        if (node.right) queue.push(node.right);  
        }
        }  
    }; // End of function levelOrder

7 months ago+ 0 comments

 def levelOrder(self,root):
    queue = [root]
    for item in queue:
        if item.left:
            queue.append(item.left)
        if item.right:
            queue.append(item.right)
    print(" ".join([str(i.data) for i in queue]))

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JavaSCript

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