数据结构笔记--常见二叉树分类及判断实现

目录

1--搜索二叉树

2--完全二叉树

3--平衡二叉树

4--满二叉树

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1--搜索二叉树

搜索二叉树的性质：左子树的节点值都比根节点小，右子树的节点值都比根节点大；

如何判断一颗二叉树是搜索二叉树？

主要思路：

        递归自底向上判断是否是一颗搜索二叉树，返回判断结果的同时，要返回对应的最小值和最大值；

#include 
#include 

struct TreeNode {
     int val;
     TreeNode *left, *right;
     TreeNode() : val(0), left(nullptr), right(nullptr) {}
     TreeNode(int x) : val(x), right(nullptr) {}
};

struct ReturnType{
    bool isBST = true;
    int max = 0;
    int min = 0;
    ReturnType(bool ib, int ma, int mi) : isBST(ib), max(ma), min(mi){}
};

class Solution {
public:
    bool isBST(TreeNode *root){
        ReturnType res = dfs(root);
        return res.isBST;
    }

    ReturnType dfs(TreeNode *root){
        // 初始化最大值为整型最小值，最小值为整型最大值
        if(root == NULL) return ReturnType(true, INT_MIN, INT_MAX); 
        ReturnType left = dfs(root->left);
        ReturnType right = dfs(root->right);

        bool isBST = true;
        if(!left.isBST || !right.isBST || left.max >= root->val || right.min <= root->val){
            isBST = false;
        }
        int min = std::min(std::min(root->val, left.min), std::min(root->val, right.min)); 
        int max = std::max(std::max(root->val, left.max), std::max(root->val, right.max));
        return ReturnType(isBST, max, min);
    }
};

int main(int argc, char *argv[]){
    TreeNode *Node1 = new TreeNode(4);
    TreeNode *Node2 = new TreeNode(2);
    TreeNode *Node3 = new TreeNode(6);
    TreeNode *Node4 = new TreeNode(1);
    TreeNode *Node5 = new TreeNode(3);
    TreeNode *Node6 = new TreeNode(5);
    TreeNode *Node7 = new TreeNode(7);


    Node1->left = Node2;
    Node1->right = Node3;
    Node2->left = Node4;
    Node2->right = Node5;
    Node3->left = Node6;
    Node3->right = Node7;


    Solution S1;
    bool res = S1.isBST(Node1);
    if (res) std::cout << "true" << std::endl;
    else std::cout << "false" << std::endl;
    return 0;
}

2--完全二叉树

如何判断一颗二叉树是完全二叉树？

主要思路：

        层次遍历二叉树的节点，当遇到第一个节点（其左右儿子不双全）进行标记，往后遇到的所有节点应均为叶子节点，当遇到一个不是叶子节点时，返回 false 表明二叉树不是完全二叉树；

#include 
#include 

struct TreeNode {
     int val;
     TreeNode *left, *right;
     TreeNode() : val(0), left(nullptr), right(nullptr) {}
     TreeNode(int x) : val(x), right(nullptr) {}
};

class Solution {
public:
    bool isCBT(TreeNode *root){
        if(root == NULL) return true;
        std::queue q;
        q.push(root);
        bool flag = false; 
        while(!q.empty()){ // 层次遍历
            TreeNode *cur = q.front();
            q.pop();
            if(cur->left != NULL) q.push(cur->left);
            if(cur->right != NULL) q.push(cur->right);
       
            if( // 标记节点后，还遇到了不是叶子节点的节点，返回false
                (flag == true && (cur->left != NULL || cur->right != NULL)) 
                || // 左儿子为空，右儿子不为空，返回false
                (cur->left == NULL && cur->right != NULL)
            ) return false;
            
            // 遇到第一个左右儿子不双全的节点进行标记
            if(cur->left == NULL || cur->right == NULL) flag = true;
        }
        return true;
    }
};

int main(int argc, char *argv[]){
    TreeNode *Node1 = new TreeNode(1);
    TreeNode *Node2 = new TreeNode(2);
    TreeNode *Node3 = new TreeNode(3);
    TreeNode *Node4 = new TreeNode(4);
    TreeNode *Node5 = new TreeNode(5);
    TreeNode *Node6 = new TreeNode(6);
    TreeNode *Node7 = new TreeNode(7);
    TreeNode *Node8 = new TreeNode(8);
    TreeNode *Node9 = new TreeNode(9);
    TreeNode *Node10 = new TreeNode(10);
    TreeNode *Node11 = new TreeNode(11);
    TreeNode *Node12 = new TreeNode(12);

    Node1->left = Node2;
    Node1->right = Node3;
    Node2->left = Node4;
    Node2->right = Node5;
    Node3->left = Node6;
    Node3->right = Node7;
    Node4->left = Node8;
    Node4->right = Node9;
    Node5->left = Node10;
    Node5->right = Node11;
    Node6->left = Node12;

    Solution S1;
    bool res = S1.isCBT(Node1);
    if (res) std::cout << "true" << std::endl;
    else std::cout << "false" << std::endl;
    return 0;
}

3--平衡二叉树

平衡二叉树要求：左子树和右子树的高度差 <= 1；

如何判断一颗二叉树是平衡二叉树？

主要思路：

        递归自底向上判断是否是一颗平衡二叉树，返回判断结果的同时，要返回对应的深度；

#include 
#include 

struct TreeNode {
     int val;
     TreeNode *left, *right;
     TreeNode() : val(0), left(nullptr), right(nullptr) {}
     TreeNode(int x) : val(x), right(nullptr) {}
};

struct ReturnType{
    bool isbalanced = true;
    int height = 0;
    ReturnType(bool ib, int h) : isbalanced(ib), height(h) {}
};

class Solution {
public:
    bool isBalanced(TreeNode *root){
        ReturnType res = dfs(root);
        return res.isbalanced;
    }

    ReturnType dfs(TreeNode *root){
        if(root == NULL) return ReturnType(true, 0);
        ReturnType left = dfs(root->left);
        ReturnType right = dfs(root->right);

        int cur_height = std::max(left.height, right.height) + 1; 
        
        bool cur_balanced = left.isbalanced && right.isbalanced 
            && std::abs(left.height - right.height) <= 1;

        return ReturnType(cur_balanced, cur_height);
    }

private:
    int height = 0;
};

int main(int argc, char *argv[]){
    TreeNode *Node1 = new TreeNode(1);
    TreeNode *Node2 = new TreeNode(2);
    TreeNode *Node3 = new TreeNode(3);
    TreeNode *Node4 = new TreeNode(4);
    TreeNode *Node5 = new TreeNode(5);
    TreeNode *Node6 = new TreeNode(6);
    TreeNode *Node7 = new TreeNode(7);
    TreeNode *Node8 = new TreeNode(8);
    TreeNode *Node9 = new TreeNode(9);
    TreeNode *Node10 = new TreeNode(10);
    TreeNode *Node11 = new TreeNode(11);
    TreeNode *Node12 = new TreeNode(12);

    Node1->left = Node2;
    Node1->right = Node3;
    Node2->left = Node4;
    Node2->right = Node5;
    Node3->left = Node6;
    Node3->right = Node7;
    Node4->left = Node8;
    Node4->right = Node9;
    Node5->left = Node10;
    Node5->right = Node11;
    Node6->left = Node12;

    Solution S1;
    bool res = S1.isBalanced(Node1);
    if (res) std::cout << "true" << std::endl;
    else std::cout << "false" << std::endl;
    return 0;
}

4--满二叉树

满二叉树的判断：节点数 = 2^深度 - 1；

如何判断一颗二叉树是平衡二叉树？

主要思路：

        递归自底向上判断是否是一颗满二叉树，返回判断结果的同时，要返回对应的深度和节点数；

#include 
#include 

struct TreeNode {
     int val;
     TreeNode *left, *right;
     TreeNode() : val(0), left(nullptr), right(nullptr) {}
     TreeNode(int x) : val(x), right(nullptr) {}
};

struct ReturnType{
    bool isFCT = true;
    int height = 0;
    int nodes = 0;
    ReturnType(bool ib, int h, int n) : isFCT(ib), height(h), nodes(n){}
};

class Solution {
public:
    bool isFCT(TreeNode *root){
        ReturnType res = dfs(root);
        return res.isFCT;
    }

    ReturnType dfs(TreeNode *root){
        if(root == NULL) return ReturnType(true, 0, 0);
        ReturnType left = dfs(root->left);
        ReturnType right = dfs(root->right);
        bool isFCT = true;
        int cur_nodes = left.nodes + right.nodes + 1;
        int cur_height = std::max(left.height, right.height) + 1;
        if(!left.isFCT || !right.isFCT || (1<left = Node2;
    Node1->right = Node3;
    Node2->left = Node4;
    Node2->right = Node5;
    Node3->left = Node6;
    Node3->right = Node7;
    Node4->left = Node8;
    Node4->right = Node9;
    Node5->left = Node10;
    Node5->right = Node11;
    Node6->left = Node12;

    Solution S1;
    bool res = S1.isFCT(Node1);
    if (res) std::cout << "true" << std::endl;
    else std::cout << "false" << std::endl;
    return 0;
}