【leetcode刷题笔记】Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1

    \

     2

    /

   3

 

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

解题：应该是很简单的一道题，纠结了好久T_T

        基本思路很简单，用栈模拟就可以了。首先根节点压栈，每次弹出栈顶元素，并把它的值存入返回值向量中。如果它的右子树不为空，就把右子树根节点压栈，左子树不为空也把左子数根节点压栈，注意一定是右左这样的顺序。

        主要纠结在两个地方：

        1. 结构体指针的初始化：

struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));

　　注意要给指针分配空间，使用malloc需要包含头文件#include<cstdlib>

     2.每次压栈以后栈顶元素就变了，所以要先把栈顶元素pop到一个变量中，再继续后续操作。

代码如下：

 1 /**

 2  * Definition for binary tree

 3  * struct TreeNode {

 4  *     int val;

 5  *     TreeNode *left;

 6  *     TreeNode *right;

 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 8  * };

 9  */

10  #include <iostream>

11  #include <stack>

12  #include <vector>

13  #include <cstdlib>

14  using namespace std;

15  struct TreeNode {

16      int val;

17      TreeNode *left;

18      TreeNode *right;

19      TreeNode(int x) : val(x), left(NULL), right(NULL) {}

20  };

21 class Solution {

22 public:

23     vector<int> preorderTraversal(TreeNode *root) {

24        stack<TreeNode*>s;

25        vector<int>ans;

26        if(root == NULL)

27             return ans;

28        s.push(root);

29        while(!s.empty()){

30             struct TreeNode* temp = s.top();

31             s.pop();

32             ans.push_back(temp->val);

33             if(temp->right != NULL)

34                 s.push(temp->right);

35             if(temp->left != NULL){

36                 s.push(temp->left);

37             }

38        }

39         return ans;

40     }

41 };

42 int main(){

43     Solution so;

44     struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));

45     struct TreeNode* root_l_c = (struct TreeNode*)malloc(sizeof(struct TreeNode));

46 

47     root->val = 2;

48     root_l_c->val = 3;

49 

50     root_l_c->left = NULL;

51     root_l_c->right = NULL;

52 

53     root->right = NULL;

54     root->left = root_l_c;

55     //cout << root->left->val<<endl;

56     so.preorderTraversal(root);

57 }

 Java版本递归解法：

 1 public class Solution {

 2     public List<Integer> preorderTraversal(TreeNode root) {

 3         List<Integer> answer = new ArrayList<>();

 4         return preorderTraversalHelper(answer, root);

 5     }

 6     public List<Integer> preorderTraversalHelper(List<Integer> answer,TreeNode root) {

 7         if(root == null)

 8             return answer;

 9         answer.add(root.val);

10         answer = preorderTraversalHelper(answer,root.left);

11         answer = preorderTraversalHelper(answer,root.right);

12         return answer;           

13     }

14 }

一个小笔记是在eclipse里面使用List或者ArrayList的时候要加上

 import java.util.ArrayList;import java.util.List;