Package g0101_0200.s0105_construct_binary_tree_from_preorder_and_inorder_traversal

Class Solution

java.lang.Object
g0101_0200.s0105_construct_binary_tree_from_preorder_and_inorder_traversal.Solution
public class Solution extends Object
105 - Construct Binary Tree from Preorder and Inorder Traversal.

Medium

Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.

Example 1:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]

Output: [3,9,20,null,null,15,7]

Example 2:

Input: preorder = [-1], inorder = [-1]

Output: [-1]

Constraints:

  • 1 <= preorder.length <= 3000
  • inorder.length == preorder.length
  • -3000 <= preorder[i], inorder[i] <= 3000
  • preorder and inorder consist of unique values.
  • Each value of inorder also appears in preorder.
  • preorder is guaranteed to be the preorder traversal of the tree.
  • inorder is guaranteed to be the inorder traversal of the tree.

To solve the “Construct Binary Tree from Preorder and Inorder Traversal” problem in Java with a Solution class, we’ll use a recursive approach. Below are the steps:

  1. Create a Solution class: Define a class named Solution to encapsulate our solution methods.

  2. Create a buildTree method: This method takes two integer arrays, preorder and inorder, as input and returns the constructed binary tree.

  3. Check for empty arrays: Check if either of the arrays preorder or inorder is empty. If so, return null, as there’s no tree to construct.

  4. Define a helper method: Define a recursive helper method build to construct the binary tree.

    • The method should take the indices representing the current subtree in both preorder and inorder.
    • The start and end indices in preorder represent the current subtree’s preorder traversal.
    • The start and end indices in inorder represent the current subtree’s inorder traversal.

  5. Base case: If the start index of preorder is greater than the end index or if the start index of inorder is greater than the end index, return null.

  6. Find the root node: The root node is the first element in the preorder array.

  7. **Find the root’s position in inorder **: Iterate through the inorder array to find the root’s position.

  8. Recursively build left and right subtrees:

    • Recursively call the build method for the left subtree with updated indices.
    • Recursively call the build method for the right subtree with updated indices.

  9. Return the root node: After constructing the left and right subtrees, return the root node.

Here’s the Java implementation:

class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder.length == 0 || inorder.length == 0) return null; // Check for empty arrays
        return build(preorder, inorder, 0, preorder.length - 1, 0, inorder.length - 1); // Construct binary tree
    }
    
    // Recursive helper method to construct binary tree
    private TreeNode build(int[] preorder, int[] inorder, int preStart, preEnd, int inStart, int inEnd) {
        if (preStart > preEnd || inStart > inEnd) return null; // Base case
        
        int rootValue = preorder[preStart]; // Root node value
        TreeNode root = new TreeNode(rootValue); // Create root node
        
        // Find root node's position in inorder array
        int rootIndex = 0;
        for (int i = inStart; i <= inEnd; i++) {
            if (inorder[i] == rootValue) {
                rootIndex = i;
                break;
            }
        }
        
        // Recursively build left and right subtrees
        root.left = build(preorder, inorder, preStart + 1, preStart + rootIndex - inStart, inStart, rootIndex - 1);
        root.right = build(preorder, inorder, preStart + rootIndex - inStart + 1, preEnd, rootIndex + 1, inEnd);
        
        return root; // Return root node
    }
    
    // TreeNode definition
    public class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;
        
        TreeNode() {}
        TreeNode(int val) { this.val = val; }
        TreeNode(int val, TreeNode left, TreeNode right) {
            this.val = val;
            this.left = left;
            this.right = right;
        }
    }
}

This implementation follows the steps outlined above and efficiently constructs the binary tree from preorder and inorder traversals in Java.

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • get

      public int get(int key)

    • buildTree

      public TreeNode buildTree(int[] preorder, int[] inorder)