day 14 二叉树part01-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_41741008/article/details/142289094

第六章二叉树part01

理论基础

需要了解二叉树的种类，存储方式，遍历方式以及二叉树的定义

递归遍历

# 前序遍历-递归-LC144_二叉树的前序遍历
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        
        def dfs(node):
            if node is None:
                return
            
            res.append(node.val)
            dfs(node.left)
            dfs(node.right)
        dfs(root)
        return res

# 中序遍历-递归-LC94_二叉树的中序遍历
class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        
        def dfs(node):
            if node is None:
                return
            
            dfs(node.left)
            res.append(node.val)
            dfs(node.right)
        dfs(root)
        return res

# 后序遍历-递归-LC145_二叉树的后序遍历
class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        
        def dfs(node):
            if node is None:
                return
            
            dfs(node.left)
            dfs(node.right)
            res.append(node.val)

        dfs(root)
        return res

代码随想录

迭代遍历（基础不好的录友，迭代法可以放过，只是看了一下思路）

迭代遍历，前序、后续和中序不一样

统一迭代（基础不好的录友，迭代法可以放过，只是看了一下思路）

这是统一迭代法的写法，如果学有余力，可以掌握一下

统一写法

层序遍历

代码随想录
10道题

102. 二叉树的层序遍历
层序遍历用的是队列 && 记录每一层的节点的个数

/**
 * 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) {
       queue<TreeNode*> que;
       vector<vector<int>> result;
       if (root !=NULL) que.push(root);
       while(!que.empty()){
        int size = que.size();
        vector<int> vec;
        for (int i = 0; i< size; i++){
            TreeNode* node = que.front();
            que.pop();
            vec.push_back(node->val);
            if (node->left) que.push(node->left);
            if (node->right) que.push(node->right);
        }
        result.push_back(vec);

       }
       return result;

    }
};

107. 二叉树的层序遍历 II
这道题比102多了一行代码

reverse(result.begin(), result.end());

class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        queue<TreeNode* > que;
        vector<vector<int>> result;
        if (root != NULL) que.push(root);
        while(!que.empty()) {
            vector<int> res;
            int size= que.size();
            for (int i=0; i<size; i++){
            TreeNode* node = que.front();
            que.pop();
            res.push_back(node->val);
            if (node->left) que.push(node->left);
            if (node->right) que.push(node->right);
        } 
        result.push_back(res);
        }
        reverse(result.begin(), result.end());
        return result;
    }
};

199. 二叉树的右视图
层序遍历，记录每层的最后一个节点即可

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        queue<TreeNode*> que;
        vector<int> res;
        if (root != NULL) que.push(root);
        while( !que.empty()){
            int size = que.size();
            for(int i = 0; i < size; i++){
                auto node = que.front();
                que.pop();
                if(node->left) que.push(node->left);
                if(node->right) que.push(node->right);
                if (i == size -1) res.push_back(node->val);
            }
        }
        return res;

    }
};

637. 二叉树的层平均值

class Solution {
public:
    vector<double> averageOfLevels(TreeNode* root) {
        vector<double> res;
        queue<TreeNode*> que;
        if (root != NULL) que.push(root);
        while(!que.empty()){
            int size = que.size();
            double sum = 0;
            for(int i=0; i<size;i++){
                auto node = que.front();
                que.pop();
                sum += node->val;
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            res.push_back(sum/size);
        }
        return res;
    }
};

429. 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) {
        vector<vector<int>> result;
        queue<Node*> que;
        if (root != NULL) que.push(root);
        while(!que.empty()){
            int size = que.size();
            vector<int> res;
            for(int i = 0; i < size; i++){
                auto node = que.front();
                que.pop();
                res.push_back(node->val);
                for(int i = 0; i< node->children.size();i++){
                    if(node->children[i]) que.push(node->children[i]);
                }
            }
            result.push_back(res);

        }
        return result;
        
    }
};

515. 在每个树行中找最大值

class Solution {
public:
    vector<int> largestValues(TreeNode* root) {
        vector<int> res;
        queue<TreeNode*> que;
        if(root != NULL) que.push(root);
        while( !que.empty()){
            int size = que.size();
            int max = INT_MIN;
            for (int i = 0;i < size; i++) {
                auto node = que.front();
                que.pop();
                if (node->val > max) max = node->val;  
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            res.push_back(max);
        }
        return res;

    }
};

515. 在每个树行中找最大值

class Solution {
public:
    Node* connect(Node* root) {
        queue<Node*> que;
        if (root != NULL) que.push(root);
        while (!que.empty()) {
            int size = que.size();
            // vector<int> vec;
            Node* nodePre;
            Node* node;
            for (int i = 0; i < size; i++) {
                if (i == 0) {
                    nodePre = que.front(); // 取出一层的头结点
                    que.pop();
                    node = nodePre;
                } else {
                    node = que.front();
                    que.pop();
                    nodePre->next = node; // 本层前一个节点next指向本节点
                    nodePre = nodePre->next;
                }
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            nodePre->next = NULL; // 本层最后一个节点指向NULL
        }
        return root;

    }
};

117. 填充每个节点的下一个右侧节点指针 II
代码和116题的完全一样

class Solution {
public:
    Node* connect(Node* root) {
        queue<Node*> que;
        if (root != NULL) que.push(root);
        while (!que.empty()) {
            int size = que.size();
            Node* nodePre;
            Node* node;
            for (int i = 0; i < size; i++) {
                if (i == 0) {
                    nodePre = que.front(); // 取出一层的头结点
                    que.pop();
                    node = nodePre;
                } else {
                    node = que.front();
                    que.pop();
                    nodePre->next = node; // 本层前一个节点next指向本节点
                    nodePre = nodePre->next;
                }
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            nodePre->next = NULL; // 本层最后一个节点指向NULL
        }
        return root;
    }
};

117. 填充每个节点的下一个右侧节点指针 II

class Solution {
public:
    int maxDepth(TreeNode* root) {
        if (root == NULL) return 0;
        int depth = 0;
        queue<TreeNode*> que;
        que.push(root);
        while(!que.empty()) {
            int size = que.size();
            depth++; // 记录深度
            for (int i = 0; i < size; i++) {
                TreeNode* node = que.front();
                que.pop();
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
        }
        return depth;
    }
};

104. 二叉树的最大深度
记录层数即可

class Solution {
public:
    int maxDepth(TreeNode* root) {
        if (root == NULL) return 0;
        int num = 0;
        queue<TreeNode*> que;
        if (root != NULL) que.push(root);
        while( !que.empty()) {
            int size = que.size();
            num += 1;
            for (int i = 0; i < size; i++){
                auto node = que.front();
                que.pop();
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }

        }
        return num;

    }
};

111. 二叉树的最小深度
注意终止的条件

class Solution {
public:
    int minDepth(TreeNode* root) {
        if (root == NULL) return 0;
        int depth = 0;
        queue<TreeNode*> que;
        if (root != NULL) que.push(root);
        while( !que.empty()) {
            int size = que.size();
            depth += 1;
            for (int i = 0;i < size; i++) {
                auto node = que.front();
                que.pop();
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
                if (!node->left &&  !node->right) return depth;
            }
        }
        return depth;

    }
};