105. Construct Binary Tree from Preorder and Inorder Traversal
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
/**
* 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 treeNode = new TreeNode(preorder[0]);
for(int i=0;i<inorder.length;i++){
if(inorder[i]==preorder[0]){
treeNode.left = buildTree(
Arrays.copyOfRange(preorder, 1, i+1),
Arrays.copyOfRange(inorder, 0 ,i)
);
treeNode.right = buildTree(
Arrays.copyOfRange(preorder, i+1, preorder.length),
Arrays.copyOfRange(inorder,i+1,preorder.length)
);
}
}
return treeNode;
}
}