105. Construct Binary Tree from Preorder and Inorder Traversal
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.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
if(preorder==null || inorder==null || preorder.length==0 || inorder.length==0){
return null;
}
TreeNode root = new TreeNode(preorder[0]);
for(int i=0;i<inorder.length;i++){
if(inorder[i] == preorder[0]){
root.left = buildTree(Arrays.copyOfRange(preorder, 1, i+1),
Arrays.copyOfRange(inorder, 0, i));
root.right = buildTree(Arrays.copyOfRange(preorder, i+1, preorder.length),
Arrays.copyOfRange(inorder, i+1, inorder.length));
}
}
return root;
}
}