dfs构建二叉树
tricky的点在于：如何确定子树在数组中对应的起点和终点位置
1、首先确认左子树长度 leftlen = inStart - mid;
则 有 preOrder顺序：
preStart + 1 -> preStart + leftlen
preStart + leftlen + 1 -> preEnd

public TreeNode buildTree(int[] preorder, int[] inorder) {
        //[][]
        //[3,1,2][1,3,2]
        //[3,4,5][3,4,5]?
        //[1,2][]?
        // [3,9,20,15,7][9,3,15,20,7]
        int n = preorder.length;
        int m = inorder.length;
        if (n  == 0 || m == 0 || (n != m)){
            return null;
        }
        return dfs (preorder, inorder,0, n -1 , 0 , m - 1);
    }
    public TreeNode dfs(int[] preorder, int[] inorder, int preStart, int preEnd, int inStart, int inEnd){
        if (preStart > preEnd || inStart > inEnd){
            return null;
        }
        int rootVal = preorder[preStart];
        TreeNode root = new TreeNode(rootVal);

        int mid = inStart;
        while (mid <= inEnd && inorder[mid] != rootVal){
            mid++;
        }

        int leftLen = mid - inStart;//1 - 0 = 1

        root.left = dfs(preorder, inorder, preStart + 1, preStart + leftLen, inStart, mid - 1);
        root.right = dfs(preorder, inorder, preStart + leftLen + 1, preEnd, mid + 1, inEnd);
        return root;
    }