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