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Binary Tree Preorder Traversal

2024-09-11 22:00:48   217人阅读

思路一：采用递归的方法，每个节点访问一遍，时间复杂度O(n)，空间复杂度O(n)

/**
 * 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:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> result;
        preorder(root, result);
        return result;
    }
    
    void preorder(TreeNode *root, vector<int> &result)
    {
        if(root != nullptr)
        {
            result.push_back(root->val);
            if(root->left != nullptr)
                preorder(root->left, result);
            if(root->right != nullptr)
                preorder(root->right, result);
        }
    }
};

思路二：非递归实现，过程中使用了栈，时间和空间复杂度同上

/**
 * 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:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> result;
        stack<TreeNode *> s;
        
        if(root != nullptr)
        {
            s.push(root);
            while(!s.empty())
            {
                TreeNode *tmp = s.top();
                s.pop();
                result.push_back(tmp->val);
                if(tmp->right != nullptr)
                    s.push(tmp->right);
                if(tmp->left != nullptr)
                    s.push(tmp->left);
            }
        }
        
        return result;
    }
};

思路三：采用morris遍历方式，时间复杂度同上，但是空间复杂度O(1)：算法理解在此

关键在于将当前子树的中的最大值（最后遍历）的右指针指向根节点，以便于左边子树访问后后回溯到右子树

具体来说：

当当前节点左指针为空时，输出对应数值，回溯

当当前节点cur左指针非空，得到其左子树最大节点：当其右节点非空，输出对应数值，同时回溯，即将其右节点指向cur；若右节点非空，置空，遍历cur右子树

/**
 * 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:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> result;
        if(root == nullptr)
            return result;
        
        TreeNode *cur = root;
        while(cur != nullptr)
        {
            if(cur->left == nullptr)
            {
                result.push_back(cur->val);
                cur = cur->right;
            }
            else
            {
                TreeNode *tmp = cur->left;
                while(tmp->right != nullptr && tmp->right != cur)
                    tmp = tmp->right;
                
                if(tmp->right == nullptr)
                {
                    result.push_back(cur->val);
                    tmp->right = cur;
                    cur = cur->left;
                }
                else
                {
                    tmp->right = nullptr;
                    cur = cur->right;
                }
            }
        }
        return result;
    }
};

 

Binary Tree Preorder Traversal


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