剑指offer之中序打印二叉树（非递归实现）

1 问题

中序打印二叉树（非递归实现），比如二叉树如下

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

中序：按左中右来打印二叉树，结果如下

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

2 代码实现

#include <iostream>
#include <stack>

using namespace std;

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

void center_print(Node *head)
{
if (head == NULL)
{
std::cout << "head is NULL" << std::endl;
return;
}
std::stack<Node *> stack;
Node *p = head;
while (p != NULL || !stack.empty())
{
while (p != NULL)
                {
    stack.push(p);
                    p = p->left;
                }
                if (!stack.empty())
                {
    p = stack.top();
                    std::cout << p->value << std::endl;
                    stack.pop();
                    p = p->right;
                }
}
} 

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;
    center_print(&head1);
    return 0;
}

3 运行结果

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