[JAVA]LeetCode105 Construct Binary Tree from Preorder and Inorder Traversal

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

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

根据前序遍历和中序遍历，构建二叉树。假设树中不存在相同的节点

解题思路：根据前序遍历的元素，每次都可以将中序遍历划分为左子树节点和右子树节点。考虑用递归的方法解题。

public class Solution
{
	 Map<Integer,Integer> inMap=new TreeMap<Integer,Integer>();//将中序的值放入map中，可以很容易找到前序遍历元素，在中序中所对应的下标
	 int preIndex=0;//记录每棵树的下标
	public TreeNode buildTree(int[] preorder, int[] inorder)
	{
	 if(preorder.length==0||inorder.length==0)return null;
	 for(int i=0;i<inorder.length;i++)
	  {
	    inMap.put(inorder[i],i);
	  }
	    return createTree(preorder,0,inorder.length-1);
	}
	 public TreeNode createTree(int[] preorder,int inStart,int inEnd)
	 {  
	    if(inStart>inEnd)return null;
	    TreeNode root=new TreeNode(preorder[preIndex++]);
	    if(inStart==inEnd)return root;
	    int inIndex=inMap.get(root.val);
	    root.left=createTree(preorder,inStart,inIndex-1);
	    root.right=createTree(preorder,inIndex+1,inEnd);
	    return root;
	 }
}