144. 二叉树的前序遍历

好巧的代码，赞叹

/**
 * 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> res;
        stack<TreeNode*> stk;
        auto pt = root;
        while(pt || stk.size())
        {
            while(pt)
            {
                res.push_back(pt->val);
                stk.push(pt);
                pt = pt->left;
            }
            if(stk.size())
            {
                pt = stk.top();
                stk.pop();
                pt = pt->right;
            }
        }
        return res;
    }
};

第二次写题

/**
 * 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> res;
        stack<TreeNode*> stk;
        auto pt = root;
        while(pt || stk.size())
        {
            while(pt)
            {
                res.push_back(pt->val);
                stk.push(pt);
                pt = pt->left;
            }
            if(stk.size())
            {
                pt = stk.top();
                stk.pop();
                pt = pt->right;
            }
        }
        return res;
    }
};

第三次写题
递归的方式

/**
 * 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 {
    vector<int> res;
    void dfs(TreeNode* root)
    {
        if(!root) return;
        res.push_back(root->val);
        dfs(root->left);
        dfs(root->right);
    }
public:
    vector<int> preorderTraversal(TreeNode* root) {
        dfs(root);
        return res;
    }
};

迭代的方式：
这个写法很有意思

/**
 * 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) {
        stack<TreeNode*> stk;
        vector<int> res;
        while(root || stk.size())
        {
            while(root)
            {
                res.push_back(root->val);
                stk.push(root);
                root = root->left;

            }
            root = stk.top();
            stk.pop();
            root = root->right;
        }
        return res;
    }
};

Morris
这里主要要注意的是，节点会走两次，第一次放入res，第二次不要放入res

/**
 * 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> res;
        while(root)
        {
            if(!root->left)
            {
                res.push_back(root->val);
                root = root->right;
            }
            else
            {
                auto tmp = root->left;
                while(tmp->right && tmp->right != root) tmp = tmp->right;
                if(tmp->right == root) 
                {
                    tmp->right = nullptr;
                    root = root->right;
                }
                else{
                    res.push_back(root->val);
                    tmp->right = root;
                    root = root->left;
                }
            }
        }
        return res;
    }
};

第三次写题

/**
 * 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> res;
        stack<TreeNode*> stk;
        while(root || stk.size())
        {
            while(root)
            {
                res.push_back(root->val);
                stk.push(root);
                root = root->left;
            }
            root = stk.top();
            stk.pop();
            root = root->right;
        }
        return res;
    }
};

Morris算法

/**
 * 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> res;
        while(root)
        {
            if(!root->left)
            {
                res.push_back(root->val);
                root = root->right;
            }
            else
            {
                TreeNode* tmp = root->left;
                while(tmp->right && tmp->right != root) tmp = tmp->right;
                if(tmp->right == root)
                {
                    tmp->right = nullptr;
                    root = root->right;
                }
                else
                {
                    res.push_back(root->val);
                    tmp->right = root;
                    root = root->left;
                }
            }
        }
        return res;
    }
};