Array used to make the BST (that my function takes as an input): [38,95,70,5,10,148,93]
As of now the the function only returns [5,10], instead of returning all elements in a sorted order. What is wrong with this approach?
function sortedArrayFromBST(tree,node,outputArray){
outputArray=outputArray||[]
// console.log(tree)
node=node||tree.root
console.log(node)
let left=node.left
console.log(left)
let right=node.right
console.log(right)
if(left ==null){
outputArray.push(node.data)
}else{
return sortedArrayFromBST(tree,left, outputArray)
}
if(right==null){
return outputArray
} else{
return sortedArrayFromBST(tree, right, outputArray)
}
}
In the heart of your recursive function you want to do something like the following:
if (node === null) // base case
return
if (node.left !== null)
sortedArrayFromBST(node.left, outputArray)
outputArray.push(node.val)
if (node.right !== null)
sortedArrayFromBST(node.right, outputArray)
return outputArray
Your conditions right now are not correct.
The return conditions you're using are wrong. Instead of returning from the left node, you need to append it's values to the output and then process the right node.
function sortedArrayFromBST(tree, node) {
if (node == null) {
return []
}
outputArray = []
let left = node.left
let right = node.right
outputArray.push(sortedArrayFromBST(tree, left))
outputArray.push(node.data)
outputArray.push(sortedArrayFromBST(tree, right))
return outputArray
}