【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：

 /**
  * Definition for binary tree
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
  #include <iostream>
  #include <stack>
  #include <vector>
  #include <cstdlib>
  using namespace std;
  struct TreeNode {
      int val;
      TreeNode *left;
      TreeNode *right;
      TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  };
 class Solution {
 public:
     vector<int> preorderTraversal(TreeNode *root) {
        stack<TreeNode*>s;
        vector<int>ans;
        if(root == NULL)
             return ans;
        s.push(root);
        while(!s.empty()){
             struct TreeNode* temp = s.top();
             s.pop();
             ans.push_back(temp->val);
             if(temp->right != NULL)
                 s.push(temp->right);
             if(temp->left != NULL){
                 s.push(temp->left);
             }
        }
         return ans;
     }
 };
 int main(){
     Solution so;
     struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
     struct TreeNode* root_l_c = (struct TreeNode*)malloc(sizeof(struct TreeNode));
 
     root->val = ;
     root_l_c->val = ;
 
     root_l_c->left = NULL;
     root_l_c->right = NULL;
 
     root->right = NULL;
     root->left = root_l_c;
     //cout << root->left->val<<endl;
     so.preorderTraversal(root);
 }

Java版本递归解法：

 public class Solution {
     public List<Integer> preorderTraversal(TreeNode root) {
         List<Integer> answer = new ArrayList<>();
         return preorderTraversalHelper(answer, root);
     }
     public List<Integer> preorderTraversalHelper(List<Integer> answer,TreeNode root) {
         if(root == null)
             return answer;
         answer.add(root.val);
         answer = preorderTraversalHelper(answer,root.left);
         answer = preorderTraversalHelper(answer,root.right);
         return answer;
     }
 }

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;