树专题

剑指 Offer II 043. 往完全二叉树添加节点 - 力扣（LeetCode） (leetcode-cn.com)

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/**
 * 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 CBTInserter {
public:

    TreeNode *root;
    int size = 0;

    void dfs(TreeNode *root) {
        if (root == nullptr) return ;
        size ++;
        dfs(root->left);
        dfs(root->right);
    }

    CBTInserter(TreeNode *root) {
        this->root = root;
        dfs(root);    
    }
    
    int insert(int v) {
        size ++;
        TreeNode* fatherNode = getFatherNode(size / 2);
        if (size % 2 == 1) {
            fatherNode->right = new TreeNode(v);
        } else {
            fatherNode->left = new TreeNode(v);
        }
        return fatherNode->val;
    }
    
    TreeNode* getFatherNode(int n) {
        if (n == 1) return root;
        TreeNode* node = getFatherNode(n / 2);
        return (n % 2 == 1) ? node->right : node->left;
    }

    TreeNode* get_root() {
        return root;
    }
};

/**
 * Your CBTInserter object will be instantiated and called as such:
 * CBTInserter* obj = new CBTInserter(root);
 * int param_1 = obj->insert(v);
 * TreeNode* param_2 = obj->get_root();
 */

[剑指 Offer II 045. 二叉树最底层最左边的值 - 力扣（LeetCode） (leetcode-cn.com)](https://leetcode-cn.com/problems/LwUNpT/)

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/**
 * 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 = -1;
    int res;

    int findBottomLeftValue(TreeNode* root) {
        dfs (root, 0);
        return res;
    }

    void dfs(TreeNode *root, int depth) {
        if (root == nullptr) return ;

        if (depth > maxDepth) {
            res = root->val;
            maxDepth = depth;
        }

        dfs(root->left, depth + 1);
        dfs(root->right, depth + 1);
    }
};

剑指 Offer II 047. 二叉树剪枝 - 力扣（LeetCode） (leetcode-cn.com)

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/**
 * 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;
        auto left = pruneTree(root->left);
        auto right = pruneTree(root->right);
        if (root->val == 0 && !left && !right) {
            return nullptr;
        }
        root->left = left;
        root->right = right;
        return root;
    }
};

剑指 Offer II 049. 从根节点到叶节点的路径数字之和 - 力扣（LeetCode） (leetcode-cn.com)

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/**
 * 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 sum;

    int sumNumbers(TreeNode* root) {
        dfs(root, 0);
        return sum;
    }

    void dfs(TreeNode *root, int path) {
        if (root == nullptr) return ;

        path = path * 10 + root->val;
        if (root->left == nullptr && root->right == nullptr) {
            sum += path;
            return ;
        }
        dfs(root->left, path);
        dfs(root->right, path);
    }
};

剑指 Offer II 050. 向下的路径节点之和 - 力扣（LeetCode） (leetcode-cn.com)

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/**
 * 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 res;

    int pathSum(TreeNode* root, int targetSum) {
        if (root == nullptr) return 0;
        dfs(root, targetSum);
        pathSum(root->left, targetSum);
        pathSum(root->right, targetSum);
        return res;
    }

    void dfs(TreeNode *root, int sum) {
        if (root == nullptr) return ;
        sum -= root->val;
        if (!sum) res ++;
        if (root->left != nullptr) dfs(root->left, sum);
        if (root->right != nullptr) dfs(root->right, sum);
    }
};

剑指 Offer II 053. 二叉搜索树中的中序后继 - 力扣（LeetCode） (leetcode-cn.com)

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *res;
    bool flag;

    void dfs(TreeNode *root, TreeNode *p) {
        if (root == NULL) return ;
        dfs(root->left, p);
        if (root->val > p->val && !flag) {
            res = root;
            flag = true;
        }
        dfs(root->right, p);
    }

    TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
        if (root == NULL) return NULL;
        dfs(root, p);
        return res;
    }
};

剑指 Offer II 054. 所有大于等于节点的值之和 - 力扣（LeetCode） (leetcode-cn.com)

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/**
 * 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 sum = 0;
    TreeNode* convertBST(TreeNode* root) {
        if (root == nullptr) return nullptr;
        convertBST(root->right);
        root->val = root->val + sum;
        sum = root->val;
        convertBST(root->left);
        return root;
    }
};
算法
#树
树专题
https://2w1nd.github.io/2022/02/18/算法/理论算法/树专题/
作者
w1nd
发布于
2022年2月18日
许可协议