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 nodestack.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 nodestack.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。