Construct Binary Tree from Preorder and Inorder Traversal

Posted 2021-02-11 beiyeqingteng

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

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

For example, given

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

Return the following binary tree:

    3
   /   9  20
    /     15   7

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public TreeNode buildTree(int[] preorder, int[] inorder) {
12         if (preorder == null || inorder == null || preorder.length == 0 || inorder.length != preorder.length) return null;
13         return buildTree(inorder, 0, preorder, 0, inorder.length);
14     }
15     
16     public TreeNode buildTree(int[] inorder, int inStart, int[] preorder, int preStart, int length) {
17         if (length == 0) return null;
18         TreeNode root = new TreeNode(preorder[preStart]);
19         int index = findIndex(inorder, root.val);
20         root.left = buildTree(inorder, inStart, preorder, preStart + 1, index - inStart);
21         root.right = buildTree(inorder, index + 1, preorder, preStart + (index - inStart) + 1, length - (index - inStart) - 1);
22         return root;    
23     }
24     
25     public int findIndex(int[] inorder, int val) {
26         for (int i = 0; i < inorder.length; i++) {
27             if (inorder[i] == val) {
28                 return i;
29             }
30         }
31         return -1;
32     }
33 }