/计算二叉树的叶节点数int LeafNode(BTNode *tree){if(tree==NULL) //空树,叶节点为0return 0;else {if((tree->lchild==NULL) && (tree->rchild==NULL))//叶节点时,计数1return 1;elsereturn (LeafNode(tree->lchild) + LeafNode(tree->rchild)); //计算左子树和右子树的叶节点数之和}}