目录

一、前序

二、中序

三、后序

---

一、前序

力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台 

 

/**
 * 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) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> v;
        stack<TreeNode*> s;
        TreeNode*cur = root;
       
        while(cur!=nullptr || !s.empty())
//cur为空表示结点为空，栈为不为空表示还有右子树没访问
        {
            //访问一棵树的开始
            //1. 访问左路结点，左路结点入栈
            while(cur!=nullptr)
            {
                v.push_back(cur->val);
                s.push(cur);
                cur = cur->left;
            }
            //依次访问左路结点的右子树
            TreeNode*tmp = s.top();
            s.pop();
            cur = tmp->right;
        }
        return v;
    }
};

二、中序

力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

 

/**
 * 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) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        stack<TreeNode*>s;//栈中存放待访问的结点
        vector<int>v;
        TreeNode*cur = root;
        while(cur!=nullptr || !s.empty())
        {
            while(cur!=nullptr)//while循环是将左路结点进行了入栈
            {
                s.push(cur);
                cur = cur->left;
            }
            TreeNode*tmp = s.top();//从栈中开始取数据，代表左子树已经访问完了
            s.pop();
            v.push_back(tmp->val);//先访问根
            cur = tmp->right;//去到右子树
        }
        return v;
    }
};

三、后序

 

/**
 * 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) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        stack<TreeNode*> s;
        vector<int> v;
        TreeNode* cur = root;
        TreeNode* prev = nullptr;//prev记录上一个访问的结点
        while(cur!=nullptr || !s.empty())
        {
            while(cur!=nullptr)
            {
                s.push(cur);
                cur = cur->left;
            }

            TreeNode* tmp = s.top();
            if(tmp->right == nullptr || tmp->right == prev)
            //如果tmp的右为空或者上一个访问的是tmp的右，此时就可以访问tmp
            {
                v.push_back(tmp->val);
                s.pop();
                prev = tmp;//更新tmp
            }
            else//tmp结点的右没有访问过，此时先访问tmp的右
            {
                cur = tmp->right;
            }
        }
        return v;
    }
};