105. Construct Binary Tree from Preorder and Inorder Traversal

Posted 2020-06-22 蜃利的阴影下

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

 

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 (M) Construct Binary Tree from Inorder and Postorder Traversal

 

 

Original version, easier to understand, but doing more copying work.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder.length == 0)
            return null;
        
        int rootValue =  preorder[0];
        TreeNode root = new TreeNode(rootValue);
        if(preorder.length == 1)
            return root;
        
        int inorderRootIndex = 0;
        for(int i = 0; i<inorder.length; ++i)
        {
            if(inorder[i] == rootValue)
            {
                inorderRootIndex = i;
                break;
            }
        }
        
        int[] preorderLeftBranch = Arrays.copyOfRange(preorder, 1, inorderRootIndex+1);
        int[] preorderRightBranch = Arrays.copyOfRange(preorder, inorderRootIndex+1, preorder.length);
        int[] inorderLeftBranch = Arrays.copyOfRange(inorder, 0, inorderRootIndex);
        int[] inorderRightBranch = Arrays.copyOfRange(inorder, inorderRootIndex+1, inorder.length);
        
        root.left = buildTree(preorderLeftBranch, inorderLeftBranch);
        root.right = buildTree(preorderRightBranch, inorderRightBranch);
        return root;
    }
}

 

 

Improved version. Avoid array copying and improved root index look up.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        Map<Integer, Integer> inMap = new HashMap<Integer, Integer>();
    
        for(int i = 0; i < inorder.length; i++) {
            inMap.put(inorder[i], i);
        }
    
        TreeNode root = buildTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1, inMap);
        return root;
    }
    
    public TreeNode buildTree(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd, 
        Map<Integer, Integer> inMap) {
        if(preStart > preEnd || inStart > inEnd) return null;
    
        TreeNode root = new TreeNode(preorder[preStart]);
        int inRoot = inMap.get(root.val);
        int numsLeft = inRoot - inStart;
    
        root.left = buildTree(preorder, preStart + 1, preStart + numsLeft, inorder, inStart, inRoot - 1, inMap);
        root.right = buildTree(preorder, preStart + numsLeft + 1, preEnd, inorder, inRoot + 1, inEnd, inMap);
    
        return root;
    }
}