代码随想录算法训练营第十三天|树的前序,中序,后序遍历、树的层序遍历

目录

前言

一、树的前序,中序,后序遍历

1、递归遍历

题目链接:

前序代码:

后序代码:

中序代码:

 2、迭代遍历(非递归算法):

前序代码:

后序代码:

中序代码:

二、树的层序遍历

题目链接:​​​​​​102. 二叉树的层序遍历 - 力扣(LeetCode)

题目链接:107. 二叉树的层序遍历 II - 力扣(LeetCode)

题目链接:199. 二叉树的右视图 - 力扣(LeetCode)

题目链接:637. 二叉树的层平均值 - 力扣(LeetCode)

题目链接:429. N 叉树的层序遍历 - 力扣(LeetCode) 

题目链接:515. 在每个树行中找最大值 - 力扣(LeetCode)

 题目链接:116. 填充每个节点的下一个右侧节点指针 - 力扣(LeetCode)

题目链接:117. 填充每个节点的下一个右侧节点指针 II - 力扣(LeetCode)

 题目链接:104. 二叉树的最大深度 - 力扣(LeetCode)

 题目链接:111. 二叉树的最小深度 - 力扣(LeetCode)


前言

树的理论基础

文章链接:代码随想录

树的前序,中序,后序遍历

文章链接:树的递归遍历

视频链接(递归遍历):LeetCode:144.前序遍历,145.后序遍历,94.中序遍历

文章链接:树的迭代遍历(非递归)

视频链接(迭代遍历):树的前序和后序遍历

视频链接(迭代遍历):树的中序遍历

树的层序遍历

文章链接:代码随想录

视频链接:LeetCode:102.二叉树的层序遍历


一、树的前序,中序,后序遍历

1、递归遍历

题目链接:

144. 二叉树的前序遍历 - 力扣(LeetCode)

145. 二叉树的后序遍历 - 力扣(LeetCode)

94. 二叉树的中序遍历 - 力扣(LeetCode)

前序代码:

/**
 * 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:
    void preOrder(vector<int>& ret, TreeNode* root)
    {
        if(root != nullptr)
        {
            ret.push_back(root->val);
            preOrder(ret, root->left);
            preOrder(ret, root->right);
        }
    }
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> ret;
        preOrder(ret, root);
        return ret;
    }
};

后序代码:

/**
 * 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:
    void postOrder(vector<int>& ret, TreeNode* root)
    {
        if(root != nullptr)
        {
            postOrder(ret, root->left);
            postOrder(ret, root->right);
            ret.push_back(root->val);
        }
    }

    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> ret;
        postOrder(ret, root);

        return ret;
    }
};

中序代码:

/**
 * 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:
    void inOrder(vector<int>& ret, TreeNode* root)
    {
        if(root == nullptr)
            return ;
        inOrder(ret, root->left);
        ret.push_back(root->val);
        inOrder(ret, root->right);;
    }
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> ret;
        inOrder(ret, root);
        return ret;
    }
};

 2、迭代遍历(非递归算法):

前序代码:

/**
 * 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<int> preorderTraversal(TreeNode* root) {
        vector<int> ret;
        stack<TreeNode*> sta;

        if(root)
            sta.push(root);
        while(!sta.empty())
        {
            TreeNode *pCur = sta.top();
            sta.pop();
            if(pCur != nullptr)
            {
                ret.push_back(pCur->val);
            }
            else
                continue;
            
            sta.push(pCur->right);
            sta.push(pCur->left);
        }

        return ret;
    }
};

后序代码:

/**
 * 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:
    void postOrder(vector<int>& ret, TreeNode* root)
    {
        if(root != nullptr)
        {
            postOrder(ret, root->left);
            postOrder(ret, root->right);
            ret.push_back(root->val);
        }
    }

    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> ret;
        postOrder(ret, root);

        return ret;
    }
};

中序代码:

/**
 * 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<int> inorderTraversal(TreeNode* root) {
        vector<int> ret;
        stack<TreeNode*> sta;
        TreeNode *pCur = root;
        
        while(pCur || !sta.empty())
        {
            if(pCur)
            {
                sta.push(pCur);
                pCur = pCur->left;
            }
            else
            {
                pCur = sta.top();
                sta.pop();
                ret.push_back(pCur->val);
                pCur = pCur->right;
            }
        }

        return ret;
    }
};

二、树的层序遍历

题目链接:​​​​​​102. 二叉树的层序遍历 - 力扣(LeetCode)

/**
 * 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>> levelOrder(TreeNode* root) {
        vector<vector<int>> ret;
        queue<TreeNode*> que;
        if(root)
            que.push(root);
        while(!que.empty())
        {
            int size = que.size();
            vector<int> vec;
            while(size--)
            {
                TreeNode *pCur = que.front();
                que.pop();
                vec.push_back(pCur->val);
                if(pCur->left)
                    que.push(pCur->left);
                if(pCur->right)
                    que.push(pCur->right);
            }
            ret.push_back(vec);
        }
        return ret;
    }
};

题目链接:107. 二叉树的层序遍历 II - 力扣(LeetCode)

/**
 * 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:
    void reverse(vector<vector<int>>& vec)
    {
        int low = 0;
        int high = vec.size() - 1;
        while(low < high)
        {
            swap(vec[low++], vec[high--]);
        }
    }
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> ret;
        queue<TreeNode *> que;

        if(root)
            que.push(root);
        
        while(!que.empty())
        {
            int size = que.size();
            vector<int> vec;
            while(size--)
            {
                TreeNode *pCur = que.front();
                que.pop();
                vec.push_back(pCur->val);
                if(pCur->left)
                {
                    que.push(pCur->left);
                }
                if(pCur->right)
                {
                    que.push(pCur->right);
                }
            }
            ret.push_back(vec);
        }
        reverse(ret);
        return ret;
    }
};

题目链接:199. 二叉树的右视图 - 力扣(LeetCode)

/**
 * 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> rightSideView(TreeNode* root) {
        vector<int> ret;
        queue<TreeNode*> que;
        if (root)
            que.push(root);

        while (!que.empty()) {
            int size = que.size();
            while (size--) {
                TreeNode* pCur = que.front();
                que.pop();
                if (pCur->left) {
                    que.push(pCur->left);
                }
                if (pCur->right) {
                    que.push(pCur->right);
                }
                if(size == 0)
                {
                    ret.push_back(pCur->val);
                }
            }

        }
        return ret;
    }
};

题目链接:637. 二叉树的层平均值 - 力扣(LeetCode)

/**
 * 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<double> averageOfLevels(TreeNode* root) {
        vector<double> ret;
        queue<TreeNode *> que;
        if(root)
            que.push(root);
        while(!que.empty())
        {
            int size = que.size();
            double sum = 0;
            TreeNode *pCur = nullptr;
            for(int i = 0; i < size; i++)
            {
                pCur = que.front();
                que.pop();
                sum += pCur->val;
                if(pCur->left)
                {
                    que.push(pCur->left);
                }
                if(pCur->right)
                {
                    que.push(pCur->right);
                }
            }
            ret.push_back(sum / size);
        }

        return ret;
    }
};

题目链接:429. N 叉树的层序遍历 - 力扣(LeetCode) 

/*
// 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) {
        if(root == nullptr) return {};
        vector<vector<int>> ret;
        queue<Node*> que;
        que.push(root);
        while(!que.empty())
        {
            int size = que.size();
            vector<int> vec;
            Node *pCur = nullptr;
            while(size--)
            {
                pCur = que.front();
                que.pop();
                vec.push_back(pCur->val);
                for(int i = 0; i < pCur->children.size(); i++)
                {
                    que.push(pCur->children[i]);
                }
            }
            ret.push_back(vec);
        }
        return ret;
    }
};

题目链接:515. 在每个树行中找最大值 - 力扣(LeetCode)

/**
 * 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) {
        if(root == nullptr) return {};
        vector<int> ret;
        queue<TreeNode*> que;
        que.push(root);
        while(!que.empty())
        {
            int size = que.size();
            TreeNode *pCur = nullptr;
            int maxNum = INT32_MIN;
            while(size--)
            {
                pCur = que.front();
                que.pop();
                if(pCur->left)
                    que.push(pCur->left);
                if(pCur->right)
                    que.push(pCur->right);
                if(pCur->val > maxNum)
                {
                    maxNum = pCur->val;
                }
            }
            ret.push_back(maxNum);
        }
        return ret;
    }
};

 题目链接:116. 填充每个节点的下一个右侧节点指针 - 力扣(LeetCode)

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
*/

class Solution {
public:
    Node* connect(Node* root) {
        if (root == nullptr)
            return {};
        queue<Node*> que;
        que.push(root);
        while (!que.empty()) {
            int size = que.size();
            Node* pCur = nullptr;
            while (size--) {
                pCur = que.front();
                que.pop();
                if (size != 0) {
                    pCur->next = que.front();

                } else {
                    pCur->next = nullptr;
                }
                if (pCur->left)
                    que.push(pCur->left);
                if (pCur->right)
                    que.push(pCur->right);
            }
        }
        return root;
    }
};

题目链接:117. 填充每个节点的下一个右侧节点指针 II - 力扣(LeetCode)

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
*/

class Solution {
public:
    Node* connect(Node* root) {
        if (root == nullptr)
            return {};
        queue<Node*> que;
        que.push(root);
        while (!que.empty()) {
            int size = que.size();
            Node* pCur = nullptr;
            while (size--) {
                pCur = que.front();
                que.pop();
                if (size != 0) {
                    pCur->next = que.front();

                } else {
                    pCur->next = nullptr;
                }
                if (pCur->left)
                    que.push(pCur->left);
                if (pCur->right)
                    que.push(pCur->right);
            }
        }
        return root;
    }
};

 题目链接:104. 二叉树的最大深度 - 力扣(LeetCode)

/**
 * 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 maxDepth(TreeNode* root) {
        if(root == nullptr) return 0;
        int layerNum = 0;
        queue<TreeNode*> que;
        que.push(root);
        while(!que.empty())
        {
            int size = que.size();
            TreeNode *pCur = nullptr;
            while(size--)
            {
                pCur = que.front();
                que.pop();
                if(pCur->left)
                    que.push(pCur->left);
                if(pCur->right)
                    que.push(pCur->right);
            }
            layerNum++;
        }

        return layerNum;
    }
};

 题目链接:111. 二叉树的最小深度 - 力扣(LeetCode)

/**
 * 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 minDepth(TreeNode* root) {
        if(root == nullptr) return 0;
        queue<TreeNode *> que;
        que.push(root);
        int layerNum = 1;
        while(!que.empty())
        {
            int size = que.size();
            TreeNode *pCur = nullptr;
            while(size--)
            {
                pCur = que.front();
                que.pop();
                if(pCur->left == nullptr && pCur->right == nullptr)
                    return layerNum;
                else
                {
                    if(pCur->left)
                        que.push(pCur->left);
                    if(pCur->right)
                        que.push(pCur->right);
                }
            }
            layerNum++;
        }
        return layerNum;
    }
};

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