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

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

public class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
if(preorder.length==0||inorder.length==0)
return null;
return BuildTreeNode(preorder,0,preorder.length-1,inorder,0,inorder.length-1); } private TreeNode BuildTreeNode(int[] preorder, int prestart, int preend, int[] inorder,
int instart, int inend) {
if(prestart>preend||instart>inend)
return null;
int temp = preorder[prestart];
TreeNode root = new TreeNode(temp);
int leftlength=0; for(int i=instart;i<=inend;i++){
if(inorder[i]!=temp)
leftlength++;
else if(inorder[i]==temp)
break;
}
root.left=BuildTreeNode(preorder, prestart+1, prestart+leftlength, inorder, instart,instart+leftlength-1);
root.right=BuildTreeNode(preorder,prestart+leftlength+1,preend,inorder,instart+leftlength+1,inend);
return root;
}
}

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