代码随想录-DAY⑩-二叉树——leetcode 144 | 94 | 145

144

思路

递归，略
迭代，用一个栈存放暂时没有写入到结果集的节点

时间复杂度：O(n)
空间复杂度：O(n)

代码

/**
 * 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) {}
 * };
 */
 //递归
class Solution {
public:
    void recur(TreeNode* root, vector<int>& res){
        if(root == nullptr){
            return;
        }
        res.push_back(root->val);
        recur(root->left, res);
        recur(root->right, res);
    }

    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        recur(root, res);
        return res;
    }
};
//迭代
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> stk;
        TreeNode* node = root;

        while (node!= nullptr || !stk.empty()){
            while (node != nullptr){
                res.emplace_back(node->val);
                stk.emplace(node);
                node = node->left;
            }
            node = stk.top();
            stk.pop();
            node = node->right;
        }
        return res;
    }
};

94

思路

递归，略
迭代，用一个栈存放暂时没有写入到结果集的节点

时间复杂度：O(n)
空间复杂度：O(n)

代码

/**
 * 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) {}
 * };
 */
 //递归
class Solution {
public:
    void recur(TreeNode* root, vector<int>& res){
        if(root == nullptr){
            return;
        }
        recur(root->left, res);
        res.push_back(root->val);
        recur(root->right, res);
    }

    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        recur(root, res);
        return res;
    }
};
//迭代
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        TreeNode* node = root;
        stack<TreeNode*> stk;

        while (node != nullptr || !stk.empty()){
            while(node != nullptr){
                stk.emplace(node);
                node = node->left;
            }
            node = stk.top();
            stk.pop();
            res.emplace_back(node->val);
            node = node->right;
        }
        return res;
    }
};

145

思路

递归，略
迭代，用一个栈存放暂时没有写入到结果集的节点

时间复杂度：O(n)
空间复杂度：O(n)

代码

/**
 * 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) {}
 * };
 */
 //递归
class Solution {
public:
    void recur(TreeNode* root, vector<int>& res){
        if(root == nullptr){
            return;
        }
        recur(root->left, res);
        recur(root->right, res);
        res.push_back(root->val);
    }

    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        recur(root, res);
        return res;
    }
};
//迭代
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        TreeNode* node = nullptr;
        stack<TreeNode*> stk;

        while (root != nullptr || !stk.empty()) {
            while (root != nullptr) {
                stk.emplace(root);
                root = root->left;
            }
            root = stk.top();
            stk.pop();
            if (root->right == nullptr || root->right == node) {
                res.emplace_back(root->val);
                node = root;
                root = nullptr;
            } else {
                stk.emplace(root);
                root = root->right;
            }
        }
        return res;
    }
};