剑指offer之先序非递归打印二叉树

1 问题

先序非递归打印二叉树

比如二叉树如下

     *                  2
     *            3            5
     *         1     4      2      3
     *      3    2 1   5  1   4  2   3  

先序原则：中左右打印节点，如果左边有节点继续要打做节点，打印会是如下结果

2
3
1
3
2
4
1
5
5
2
1
4
3
2
3

2 分析

我们可以用stack,先进后出，我们先push头结点，然后再push它的右节点和左节点，依次类推

3 代码实现

#include <iostream>
#include <stack>

using namespace std;

typedef struct Node
{
    int value;
    struct Node* left;
    struct Node* right;
} Node;

void start_print(Node *head)
{
if (head == NULL)
{
std::cout << "head is NULL" << std::endl;
return;
}
std::stack<Node *> stack;
stack.push(head);
while (stack.size())
{
Node *node = stack.top();
std::cout << node->value << std::endl;
//do not remember pop node
                stack.pop();
if (node->right)
{
stack.push(node->right);
}
if (node->left)
{
stack.push(node->left);
}
}
}

int main()
{
    /*                  2
     *            3            5
     *         1     4      2      3
     *      3    2 1   5  1   4  2   3
     */
    Node head1, node1, node2, node3, node4, node5, node6;
    Node node7, node8, node9, node10, node11, node12, node13, node14;
    head1.value = 2;
    node1.value = 3;
    node2.value = 5;
    node3.value = 1;
    node4.value = 4;
    node5.value = 2;
    node6.value = 3;
    node7.value = 3;
    node8.value = 2;
    node9.value = 1;
    node10.value = 5;
    node11.value = 1;
    node12.value = 4;
    node13.value = 2;
    node14.value = 3;

    head1.left = &node1;
    head1.right = &node2;

    node1.left = &node3;
    node1.right = &node4;

    node2.left = &node5;
    node2.right = &node6;

    node3.left = &node7;
    node3.right = &node8;
    node4.left = &node9;
    node4.right = &node10;
    node5.left = &node11;
    node5.right = &node12;
    node6.left = &node13;
    node6.right = &node14;

    node7.left = NULL;
    node7.right = NULL;
    node8.left = NULL;
    node8.right =  NULL;
    node9.left = NULL;
    node9.right = NULL;
    node10.left = NULL;
    node10.right = NULL;
    node11.left = NULL;
    node11.right = NULL;
    node12.left = NULL;
    node12.right = NULL;
    node13.left = NULL;
    node13.right = NULL;
    node14.left = NULL;
    node14.right = NULL;
    start_print(&head1);
    return 0;
}

4 运行结果

2
3
1
3
2
4
1
5
5
2
1
4
3
2
3

5 总结

void start_print(Node *head)
{
if (head == NULL)
{
std::cout << "head is NULL" << std::endl;
return;
}
std::stack<Node *> stack;
stack.push(head);
while (stack.size())
{
Node *node = stack.top();
std::cout << node->value << std::endl;
if (node->right)
{
stack.push(node->right);
}
if (node->left)
{
stack.push(node->left);
}
//do not remember pop node
                stack.pop();
}
}

一开始我出现了2个问题

问题1：没有调用stack.pop()函数，导致死循环。

问题2：我把那个stack.pop()写出上面的那个位置，很明显这里是栈，如果node->right或者node->left加到栈里面去了，这个时候再弹出来肯定不是我想要的效果，受之前使用queue的影响，因为pop()函数放哪里都行，想下如果是queue的话，因为是先进先出，所以如果node->right或者node->left加到队列里面去了，再pop()依然是弹出的最顶上的位置，所以不受位置限制。

小结：要记得使用pop()函数弹出来，然后stack调用pop()函数有位置限制，但是queue没有限制。