165、从中序与后序遍历序列构造二叉树

题目描述：

和上面一道题目类似，这里因为我在主函数传入的是length-1，导致调bug半天，无奈

class Solution {
    	public TreeNode buildTree(int[] inorder, int[] postorder) {
		if(null == inorder || inorder.length == 0){
			return null;
		}
//		找到根节点
		int rootnum = postorder[postorder.length - 1];
		int i = 0;
		for (; i < inorder.length; i++) {
			if(inorder[i] == rootnum){
				break;
			}
		}
            
		TreeNode root = new TreeNode(rootnum);
//		找到左子树的节点
           
            System.out.println("start" + i);
		root.left = getleft(inorder, postorder,0, i, i - 1);
//		右子树的节点下标
		root.right = getleft(inorder, postorder, i + 1, postorder.length, postorder.length - 2);
		return root;
    }
	public TreeNode getleft(int []inorder,int[] postorder,int start,int end,int rootnum){
		if(start >= end){
			return null;
		}
		TreeNode root = new TreeNode(postorder[rootnum]);
		int i = start;
		for (; i < end; i++) {
			if(inorder[i] == postorder[rootnum]){
				break;
			}
		}
//		找到左子树的节点
		root.left = getleft(inorder, postorder,start, i ,rootnum - end + i);
//		右子树的节点下标
		root.right = getleft(inorder, postorder, i + 1, end, rootnum - 1);
		return root;
		
	}
}