lintcode-medium-Construct Binary Tree from Preorder and Inorder Traversal

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

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) {
// write your code here

if(preorder == null || preorder.length == 0)
return null;

int size = preorder.length;

TreeNode root = build(preorder, 0, size - 1, inorder, 0, size - 1);

return root;
}

public TreeNode build(int[] preorder, int pre_start, int pre_end, int[] inorder, int in_start, int in_end){

if(pre_start > pre_end || in_start > in_end)
return null;

TreeNode root = new TreeNode(preorder[pre_start]);

int k = in_start;
for(; k <= in_end; k++){
if(inorder[k] == preorder[pre_start])
break;
}

TreeNode left = build(preorder, pre_start + 1, pre_start + k - in_start, inorder, in_start, k - 1);
TreeNode right = build(preorder, pre_start + k - in_start + 1, pre_end, inorder, k + 1, in_end);

root.left = left;
root.right = right;

return root;
}

}