Objective: – Given a inorder and level order traversal, construct a binary tree from that.
Input: Inorder and level order traversal
Approach:
int[] inOrder = { 4, 2, 5, 1, 6, 3, 7 };
int[] levelOrder = { 1, 2, 3, 4, 5, 6, 7 };
- First element in the levelorder [] will be the root of the tree, here it is 1.
- Now the search element 1 in inorder[], say you find it at position i, once you find it, make note of elements which are left to i (this will construct the leftsubtree) and elements which are right to i ( this will construct the rightSubtree).
- Suppose in previous step, there are X number of elements which are left of ‘i’ (which will construct the leftsubtree), but these X elements will not be in the consecutive in levelorder[] so we will extract these elements from levelorder[] by maintaining their sequence and store it in an array say newLeftLevel[].
- Similarly if there are Y number of elements which are right of ‘i’ (which will construct the rightsubtree), but these Y elements will not be in the consecutive in levelorder[] so we will extract these elements from levelorder[] by maintaining their sequence and store it in an array say newRightLevel[].
- From previous two steps construct the left and right subtree and link it to root.left and root.right respectively by making recursive calls using newLeftLevel[] and newRightLevel[].
- See the picture for better explanation.
Complete Code:
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| public class InorderLevelOrderToTree { | |
| public Node makeBTree(int[] inorder, int[] levelOrder, int iStart, int iEnd) { | |
| if (iStart > iEnd) { | |
| return null; | |
| } | |
| int rootVal = levelOrder[0]; | |
| Node root = new Node(rootVal); | |
| if (iStart == iEnd) { | |
| return root; | |
| } | |
| int index = findIndex(inorder, rootVal, iStart, iEnd); | |
| int[] newleftLevel = newLevelOrder(inorder, levelOrder, iStart, | |
| index – 1); | |
| int[] newrighttLevel = newLevelOrder(inorder, levelOrder, index + 1, | |
| iEnd); | |
| root.left = makeBTree(inorder, newleftLevel, iStart, index – 1); | |
| root.right = makeBTree(inorder, newrighttLevel, index + 1, iEnd); | |
| return root; | |
| } | |
| public int[] newLevelOrder(int[] inorder, int[] levelOrder, int iStart, | |
| int iEnd) { | |
| int[] newlevel = new int[iEnd – iStart + 1]; | |
| int x = 0; | |
| for (int i = 0; i < levelOrder.length; i++) { | |
| if (findIndex(inorder, levelOrder[i], iStart, iEnd) != –1) { | |
| newlevel[x] = levelOrder[i]; | |
| x++; | |
| } | |
| } | |
| return newlevel; | |
| } | |
| public int findIndex(int[] inorder, int value, int iStart, int iEnd) { | |
| int x = –1; | |
| for (int i = iStart; i <= iEnd; i++) { | |
| if (value == inorder[i]) { | |
| x = i; | |
| } | |
| } | |
| return x; | |
| } | |
| public void printINORDER(Node root) { | |
| if (root != null) { | |
| printINORDER(root.left); | |
| System.out.print(" " + root.data); | |
| printINORDER(root.right); | |
| } | |
| } | |
| public static void main(String args[]) { | |
| int[] inOrder = { 4, 2, 5, 1, 6, 3, 7 }; | |
| int[] levelOrder = { 1, 2, 3, 4, 5, 6, 7 }; | |
| InorderLevelOrderToTree i = new InorderLevelOrderToTree(); | |
| Node x = i.makeBTree(inOrder, levelOrder, 0, inOrder.length – 1); | |
| System.out.println("inorder traversal of constructed tree : "); | |
| i.printINORDER(x); | |
| } | |
| } | |
| class Node { | |
| int data; | |
| Node left; | |
| Node right; | |
| public Node(int data) { | |
| this.data = data; | |
| left = null; | |
| right = null; | |
| } | |
| } |
Output:
inorder traversal of constructed tree : 4 2 5 1 6 3 7
Great explanation! Thank you 🙂
Thanks Holden 🙂
Nicely explained. What is the time complexity of program?
Worst case it would be O(n^3) . extracting the elements in level order[] would be O(n^2) and one traversal for inorder[]. I will update the post. Thanks Jayesh.
I have one question. might be silly but would like to understand how Worst case time complexity would
be O(n^3)? what I am getting is O(n^2). one traversal for inorder[] doesn’t make it O(n^3).
Inorder traversal takes O(n) time and for each node traversal we are constructing new levelorder[] which takes O(n^2) time so total it takes O(n^3). Please correct me if I m wrong.
Any better algorithm available?