剑指Offer-第6天 搜索与回溯算法(简单)
第一题
题目链接:https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-lcof/
个人题解:BFS即可
代码:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector levelOrder(TreeNode* root) {
vector res;
if(!root) return {};
queue q;
q.push(root);
while(!q.empty()){
auto t=q.front();
q.pop();
res.push_back(t->val);
if(t->left) q.push(t->left);
if(t->right) q.push(t->right);
}
return res;
}
};
运行截图:
第二题
题目链接:https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-ii-lcof/
个人题解:开二维数组,每次填入空的一维数组,重复第一题的操作即可。
代码:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector> levelOrder(TreeNode* root) {
vector> res;
if(!root) return res;
queue q;
q.push(root);
while(!q.empty())
{
int n=q.size();
res.push_back(vector());
for(int i=1;i<=n;i++)
{
auto t=q.front();
q.pop();
res.back().push_back(t->val);
if(t->left) q.push(t->left);
if(t->right) q.push(t->right);
}
}
return res;
}
};
运行截图:
第三题
题目链接:https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-iii-lcof/
个人题解:设置一个 \(idx\),偶数的时候反转数组即可,其他操作和第二题一样。
代码:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector> levelOrder(TreeNode* root) {
vector> res;
if(!root) return res;
queue q;
q.push(root);
int i=1;
while(q.size()){
int len=q.size();
vector lev;
for(int j=0;jval);
if(t->left) q.push(t->left);
if(t->right) q.push(t->right);
}
if(i%2==0) reverse(lev.begin(),lev.end());
res.push_back(lev); i++;
}
return res;
}
};
运行截图: