leetcode算法之队列

在leetcode中，队列一般都是搭配BFS，即宽度优先搜索算法进行使用

1.N叉树的层序遍历

N叉树的层序遍历

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> children;

    Node() {}

    Node(int _val) {
        val = _val;
    }

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        //使用队列
        vector<vector<int>> ret;
        if(root == nullptr) return ret;
        queue<Node*> q;
        q.push(root);
        while(q.size())
        {
            int sz = q.size();
            vector<int> tmp;
            for(int i=0;i<sz;i++)
            {
                Node* t = q.front();
                q.pop();
                tmp.push_back(t->val);
                for(auto children : t->children)
                {
                    if(children)
                    {
                        q.push(children);
                    }
                }
            }
            ret.push_back(tmp);
        }
        return ret;
    }
};

2.二叉树的锯齿形层序遍历

二叉树的锯齿形层序遍历

/**
 * 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<vector<int>> zigzagLevelOrder(TreeNode* root) {
        //增加一个标志位，偶数行的时候将此行的值逆序即可
        vector<vector<int>> ret;
        if(root == nullptr) return ret;
        queue<TreeNode*> q;
        int level = 1;//标志位
        q.push(root);
        while(q.size())
        {
            int sz = q.size();
            vector<int> tmp;
            for(int i=0;i<sz;i++)
            {
                TreeNode* t = q.front();
                q.pop();
                tmp.push_back(t->val);
                if(t->left) q.push(t->left);
                if(t->right) q.push(t->right);
            }
            //判断是否需要逆序
            if(level%2 == 0) reverse(tmp.begin(),tmp.end());
            ret.push_back(tmp);
            level++;
        }
        return ret;
    }
};

3.二叉树的最大宽度

二叉树的最大宽度

/**
 * 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:
    int widthOfBinaryTree(TreeNode* root) {
        vector<pair<TreeNode*,unsigned int>> q;//使用数组来模拟队列
        q.push_back({root,1});
        unsigned int ret = 0;
        while(q.size())
        {
            //先计算此行的最大宽度
            auto&[x1,y1]  = q[0];
            auto&[x2,y2] = q.back();
            ret = max(ret,y2-y1+1);
            //再让下一行元素入队列
            //由于数组头删时间复杂度太高，我们可以再开一个数组，再覆盖
            vector<pair<TreeNode*,unsigned int>> tmp;
            for(auto&[x,y]:q)
            {
                if(x->left) tmp.push_back({x->left,2*y});
                if(x->right) tmp.push_back({x->right,2*y+1});
            }
            q = tmp;
        }
        return ret;
    }
};

4.在每个树行中找最大值

在每个树行中找最大值

/**
 * 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<int> largestValues(TreeNode* root) {
        vector<int> ret;
        if(root == nullptr) return ret;
        queue<TreeNode*> q;
        q.push(root);
        while(q.size())
        {
            int tmp = INT_MIN;
            int sz = q.size();
            for(int i=0;i<sz;i++)
            {
                TreeNode* t = q.front();
                q.pop();
                tmp = max(tmp,t->val);
                if(t->left) q.push(t->left);
                if(t->right) q.push(t->right);
            }
            ret.push_back(tmp);
        }
        return ret;
    }
};