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Binary Tree Inorder Traversal

2024-09-11 22:09:26   208人阅读

思路一：递归版本

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

思路二：非递归版本，一般二叉树的中序遍历需要记录节点出栈的次数，在中序遍历中，当节点第二次出栈时才输出对应值，这里巧妙的使用一个额外的指针实现了这个功能

以上两种方法的时间和空间复杂度都是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> inorderTraversal(TreeNode* root) {
        vector<int> result;
        stack<TreeNode *> s;
        
        TreeNode *p = root;
        while(!s.empty() || p != nullptr)
        {
            if(p != nullptr)
            {
                s.push(p);
                p = p->left;
            }
            else
            {
                TreeNode *tmp = s.top();
                result.push_back(tmp->val);
                s.pop();
                p = tmp->right;
            }
        }
        
        return result;
    }
};

思路三：使用Morris方法中序遍历，空间复杂度是O(1)

 

/**
 * 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> inorderTraversal(TreeNode* root) {
        vector<int> 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)
                {
                    tmp->right = cur;
                    cur = cur->left;
                }
                else
                {
                    result.push_back(cur->val);
                    cur = cur->right;
                    tmp->right = nullptr;
                }
            }
        }
    
        return result;
    }
};

 

Binary Tree Inorder Traversal


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