Question:

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

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

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder.length<1)
    		return null;
       return subBuild(preorder,0,preorder.length-1,inorder,0,inorder.length-1); 
    }

	private TreeNode subBuild(int[] preorder, int prestart, int preend, int[] inorder,
			int instart, int inend) {
		int root = preorder[prestart];
		TreeNode tn = new TreeNode(root);
		if(prestart==preend) return tn;
		int index = instart;
		while(inorder[index]!=root){
			index++;
		}
		int leftlen = index-instart-1;
		int rightlen = inend-index-1;
		if(index!=instart){
			tn.left=subBuild(preorder, prestart+1, leftlen+prestart+1, inorder, instart, index-1);
		}else {
			tn.left=null;
		}
		if(index!=inend){
			tn.right = subBuild(preorder, preend-rightlen, preend, inorder, index+1, inend);
		}else {
			tn.right = null;
		}
		return tn;
	}
}

leetcode JAVA Construct Binary Tree from Preorder and Inorder Traversal 难度系数3 3.37