从前序与中序遍历序列构造二叉树（中等）

先从前序遍历列表取出第一个元素，这个元素就是根节点，然后从中序遍历中找到这个根节点，节点左侧就是该节点的左子树的节点集合，右侧就是该节点的右侧节点集合，然后递归构建左右子树。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private Map<Integer, Integer> indexMap;

    public TreeNode myBuildTree(int[] preorder,int[] inorder,int preorder_left,int preorder_right,int inorder_left,int inorder_right){
        if(preorder_left>preorder_right){
            return null;
        }
        int preorder_root=preorder_left;
        int inorder_root=indexMap.get(preorder[preorder_root]);
        TreeNode root=new TreeNode(preorder[preorder_root]);
        int size_left_subtree=inorder_root-inorder_left;
        root.left=myBuildTree(preorder,inorder,preorder_left+1,preorder_left+size_left_subtree,inorder_left,inorder_root-1);
        root.right=myBuildTree(preorder,inorder,preorder_left+size_left_subtree+1,preorder_right,inorder_root+1,inorder_right);
        return root;
        
    }
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        int n=inorder.length;
        indexMap=new HashMap<Integer,Integer>();
        for(int i=0;i<n;i++){
            indexMap.put(inorder[i],i);
        }
        return myBuildTree(preorder,inorder,0,n-1,0,n-1);
       
    }
}