LeetCode 114. 二叉树展开为链表（/）

20230317 顺利通过
20230319 顺利通过
20230403 顺利通过
原题解

题目

约束

题解

方法一

递归

class Solution {
public:
    void flatten(TreeNode* root) {
        vector<TreeNode*> l;
        preorderTraversal(root, l);
        int n = l.size();
        for (int i = 1; i < n; i++) {
            TreeNode *prev = l.at(i - 1), *curr = l.at(i);
            prev->left = nullptr;
            prev->right = curr;
        }
    }

    void preorderTraversal(TreeNode* root, vector<TreeNode*> &l) {
        if (root != NULL) {
            l.push_back(root);
            preorderTraversal(root->left, l);
            preorderTraversal(root->right, l);
        }
    }
};

迭代

class Solution {
public:
    void flatten(TreeNode* root) {
        auto v = vector<TreeNode*>();
        auto stk = stack<TreeNode*>();
        TreeNode *node = root;
        while (node != nullptr || !stk.empty()) {
            while (node != nullptr) {
                v.push_back(node);
                stk.push(node);
                node = node->left;
            }
            node = stk.top(); stk.pop();
            node = node->right;
        }
        int size = v.size();
        for (int i = 1; i < size; i++) {
            auto prev = v.at(i - 1), curr = v.at(i);
            prev->left = nullptr;
            prev->right = curr;
        }
    }
};

方法二

class Solution {
public:
    void flatten(TreeNode* root) {
        if (root == nullptr) {
            return;
        }
        auto stk = stack<TreeNode*>();
        stk.push(root);
        TreeNode *prev = nullptr;
        while (!stk.empty()) {
            TreeNode *curr = stk.top(); stk.pop();
            if (prev != nullptr) {
                prev->left = nullptr;
                prev->right = curr;
            }
            TreeNode *left = curr->left, *right = curr->right;
            if (right != nullptr) {
                stk.push(right);
            }
            if (left != nullptr) {
                stk.push(left);
            }
            prev = curr;
        }
    }
};

解法三

class Solution {
public:
    void flatten(TreeNode* root) {
        TreeNode *curr = root;
        while (curr != nullptr) {
            if (curr->left != nullptr) {
                auto next = curr->left;
                auto predecessor = next;
                while (predecessor->right != nullptr) {
                    predecessor = predecessor->right;
                }
                predecessor->right = curr->right;
                curr->left = nullptr;
                curr->right = next;
            }
            curr = curr->right;
        }
    }
};