DFS：二叉树中的深搜

news2025/1/14 18:05:29

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文章目录

前言

一. 深搜的总结

二. 二叉树的深搜题目

2.1 计算布尔二叉树的值

2.2 求根节点到叶节点的数字之和

2.3 二叉树剪枝

2.4 验证二叉搜索树

2.5 二叉搜索树中第k小的节点

2.6 二叉树的所有路径

总

前言

本篇详细介绍了二叉树中的深搜，让使用者了解二叉树中的深搜，而不是仅仅停留在表面，文章可能出现错误，如有请在评论区指正，让我们一起交流，共同进步！

一. 深搜的总结

二. 二叉树的深搜题目

2.1 计算布尔二叉树的值

2331. 计算布尔二叉树的值 - 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:
    bool evaluateTree(TreeNode* root) {
        if(root->left == nullptr)   return root->val == 0 ? false : true;
        bool left = evaluateTree(root->left);
        bool right = evaluateTree(root->right);
        return root->val == 2 ? left|right : left&right;
    }
};

2.2 求根节点到叶节点的数字之和

129. 求根节点到叶节点数字之和 - 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 dfs(TreeNode* root,int presum)
    {
        presum = presum*10+ root->val;
        if(root->left==nullptr&&root->right==nullptr)   return presum;
        int ret = 0;
        if(root->left)  ret+= dfs(root->left,presum);
        if(root->right) ret+= dfs(root->right,presum);
        return ret;
    }
    int sumNumbers(TreeNode* root) {
        return dfs(root,0);
    }
};

2.3 二叉树剪枝

814. 二叉树剪枝 - 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:
    TreeNode* pruneTree(TreeNode* root) {
        if(root == nullptr) return nullptr;
        root->left = pruneTree(root->left);
        root->right = pruneTree(root->right);
        if(root->left == nullptr && root->right == nullptr && root->val == 0)
        {
            //delete root; //可加可不加，防止内存泄漏
            root = nullptr;
        }
        return root;
    }
};

2.4 验证二叉搜索树

98. 验证二叉搜索树 - 力扣（LeetCode）

未剪枝版本：

/**
 * 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 {
    long prev = LONG_MIN;
public:
    bool isValidBST(TreeNode* root) {
        if(root == nullptr) return true;
        bool left = isValidBST(root->left);
        bool cur = false; //判断当前位置
        if(root->val > prev)
            cur = true;
        prev = root->val;
        bool right = isValidBST(root->right);
        return left && right && cur;
    }
};

剪枝：

/**
 * 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 {
    long prev = LONG_MIN;
public:
    bool isValidBST(TreeNode* root) {
        if(root == nullptr) return true;
        bool left = isValidBST(root->left);
        //剪枝
        if(left == false)   return false;

        bool cur = false; //判断当前位置
        if(root->val > prev)
            cur = true;
        //剪枝
        if(cur == false)   return false;

        prev = root->val;
        bool right = isValidBST(root->right);
        return left && right && cur;
    }
};

2.5 二叉搜索树中第k小的节点

230. 二叉搜索树中第K小的元素 - 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 {
    int count;
    int ret;
public:
    int kthSmallest(TreeNode* root, int k) {
        count = k;
        dfs(root);
        return ret;
    }

    void dfs(TreeNode* root)
    {
        if(root == nullptr || count == 0) return;
        dfs(root->left);
        count--;
        if(count == 0)  ret = root->val;
        dfs(root->right);
    }
};

2.6 二叉树的所有路径

257. 二叉树的所有路径 - 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 {
    vector<string> ret;
    string path;
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        dfs(root,path);
        return ret;
    }

    void dfs(TreeNode* root,string path)
    {
        path+= to_string(root->val);
        if(root->left == nullptr && root->right == nullptr) 
        {
            ret.push_back(path);
            return;
        }
        path += "->";
        if(root->left)  dfs(root->left,path);
        if(root->right) dfs(root->right,path);
    }
};