自下而上、从右往左层次遍历

试给出二叉树的自下而上、从右到左的层次遍历算法 分析： 我们只需要在层次遍历的基础上加入栈的使用，我们每次出队后的数据将其入栈，队列空了时，再去依次访问栈中元素，即可达到要求 代码如下：

struct biTree {
	char data;
	struct biTree *lchild;
	struct biTree *rchild;
};
struct Squeue {
	biTree *arr;
	int front, rear;
};
struct Stack {
	biTree *arr;
	int len;
	int top;
};
#include <stdio.h>
#include <stdlib.h>
void levelOrder2(biTree *T, Squeue *sq, int maxSize) {
	struct Stack *s = (struct Stack *)malloc(sizeof(struct Stack));
	struct biTree *p = T;
	struct biTree *r = (struct biTree *)malloc(sizeof(struct biTree));
	bool enQueueS(Squeue *, biTree *, int);
	bool isEmpty(Squeue *);
	bool deQueueS(Squeue *, biTree *, int);

	Stack *createStack(int);
	bool pushS(Stack *,biTree *);
	bool empty(Stack *);
	biTree *top(Stack *);
	bool pop(Stack *);

	s = createStack(maxSize);
	enQueueS(sq, p, maxSize);
	while (!isEmpty(sq)) {
		deQueueS(sq, r, maxSize);
		pushS(s,r);
		if (r->lchild)enQueueS(sq, r->lchild, maxSize);
		if (r->rchild)enQueueS(sq, r->rchild, maxSize);
	}
	while (!empty(s)) {
		r = top(s);
		printf("%c ",r->data);
		pop(s);
	}

}
int main() {
	int count = 0;
	struct biTree *T = (struct biTree *)malloc(sizeof(struct biTree));
	struct Squeue *sq = (struct Squeue *)malloc(sizeof(struct Squeue));

	biTree *create(biTree *);
	void nodeNum(biTree *, int *);

	Squeue *createQueue(int);
	T = create(T);//创建一颗二叉树
	nodeNum(T, &count);//统计二叉树节点个数
	sq = createQueue(count);

	levelOrder2(T, sq, count);
	return 0;
}

付出不在于多少，只要值得就好。