Construct Binary Tree from Preorder and Inorder Traversal

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

 Notice

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

Example

Given in-order 

[1,2,3]

 and
pre-order 

[2,1,3]

, return a tree:

2
/ \
1   3

/**
* Definition of TreeNode:
* public class TreeNode {
*     public int val;
*     public TreeNode left, right;
*     public TreeNode(int val) {
*         this.val = val;
*         this.left = this.right = null;
*     }
* }
*/

public class Solution {
/**
*@param preorder : A list of integers that preorder traversal of a tree
*@param inorder : A list of integers that inorder traversal of a tree
*@return : Root of a tree
*/
public TreeNode buildTree(int[] preorder, int[] inorder) {
if(preorder.length != inorder.length) {
return null;
}
return construct(inorder, 0, inorder.length - 1, preorder, 0, preorder.length - 1);
}

private TreeNode construct(int[] inorder, int inStart, int inEnd, int[] preorder, int preStart, int preEnd) {
if(inStart > inEnd || preStart > preEnd) {
return null;
}
//use preorder array to construct the root node
TreeNode root = new TreeNode(preorder[preStart]);
int position = 0;
//use the position of root in inorder array to decide the length of left subtree and right subtree
for(int i = 0; i < inorder.length; i++) {
if(inorder[i] == preorder[preStart]) {
position = i;
}
}
int leftLength = position - inStart;
root.left = construct(inorder, inStart, position - 1, preorder, preStart + 1, preStart + leftLength);
root.right = construct(inorder, position + 1, inEnd, preorder, preStart + 1 + leftLength, preEnd);
return root;
}
}