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Construct a binary tree from given Inorder and Level Order Traversal
Objective: - Given an 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 };
- The 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 the previous step, there is X number of elements which are left of 'i' (which will construct the left subtree), 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 right subtree), 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 the 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 a better explanation.
[caption id="attachment_1869" align="aligncenter" width="400"] Construct-a-binary-tree-from-given-Inorder-and-Level-Order-Traversal[/caption]
Code:
Output:
inorder traversal of constructed tree : 4 2 5 1 6 3 7
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