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

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

给出树的中序和先序遍历 要求返回树  中序结果｛1，2，5，4，3，6｝ 先序｛4，2，1，5，3，6｝根据先序可以确定根节点为preorder[0] 从中序中可以以根节点为界分出左右子树节点 进行递归赋值 代码如下：

public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
       if(preorder.length==0&&inorder.length==0){
			return null;
		}
		TreeNode root=new TreeNode(preorder[0]);
		int begindex=0;
		for(int i=0;i<inorder.length;i++){
			if(inorder[i]==preorder[0]){
				begindex=i;
				break;
			}
		}
		int leftlen=begindex;
	    int rightlen=inorder.length-begindex-1;
	    int[] leftpre=new int[leftlen];
	    int[] leftino=new int[leftlen];
		for(int i=0;i<begindex;i++){
			leftpre[i]=preorder[i+1];
			leftino[i]=inorder[i];			
		}
		root.left=buildTree(leftpre, leftino);
		int[] rightpre=new int[rightlen];
		int[] rightino=new int[rightlen];
		for(int i=0;i<rightlen;i++){
			rightpre[i]=preorder[i+begindex+1];
			rightino[i]=inorder[i+begindex+1];			
		}
		root.right=buildTree(rightpre, rightino);
		return root; 
    }
}

Construct Binary Tree from Preorder and Inorder Traversal