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




 

作者：chen.yu
深信服三年半工作经验，目前就职游戏厂商，希望能和大家交流和学习,
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