LeetCode106 根据中序和后序遍历构建二叉树

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

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

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:

3

/
9 20
/
15 7

public TreeNode buildTree(int[] inorder, int[] postorder) {
        int m = inorder.length - 1;
        int n = postorder.length - 1;
        return buildCore(inorder, 0, m, postorder, 0, n);
    }
    public TreeNode buildCore(int[] inorder, int startIn, int endIn, int[] postorder, int startPost, int endPost){
        if(startPost > endPost || startIn > endIn) return null;
        TreeNode root = new TreeNode(postorder[endPost]);
        if(startPost == endPost) return root;
        int i;
        for(i = startIn; i <= endIn; i++){
            if(inorder[i] == root.val) break;
        }
        int leftlen = i - startIn;
        if(leftlen > 0){
            root.left = buildCore(inorder, startIn, i - 1, postorder, startPost, startPost + leftlen - 1);
        }
        if(leftlen < endIn - startIn){
            root.right = buildCore(inorder, i+1, endIn, postorder, startPost +leftlen , endPost - 1);
        }
        return root;
    }