https://leetcode.cn/problems/binary-tree-preorder-traversal/submissions/

/**
 * 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) {
    stack<TreeNode*> st;
    vector<int> v;
    TreeNode* cur=root;
    while(cur || !st.empty())
    {
        //左子树
        while(cur)
        {
            st.push(cur);
            v.push_back(cur->val);

            cur=cur->left;
        }

        // 2、左路节点的右子树需要访问
        TreeNode* top=st.top();
        st.pop();

        //右子树
        cur=top->right; // 子问题访问右子树
    }

    return v;
    }
};



https://leetcode.cn/problems/binary-tree-inorder-traversal/submissions/

/**
 * 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*> st;
		vector<int> v;
		TreeNode* cur = root;
		while (cur || !st.empty())
		{
			// 开始访问一棵树
			//1、左路节点
			while (cur)
			{
				st.push(cur);

				cur = cur->left;
			}


			// 2、左路节点的右子树需要访问. 根要存入。
			TreeNode* top = st.top();
			st.pop();

            v.push_back(top->val);
			cur = top->right; // 子问题访问右子树
		}

		return v;
	}
};

https://leetcode.cn/problems/binary-tree-postorder-traversal/submissions/


/**
 * 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*> st;
    vector<int> v;

    TreeNode* cur=root, *prev=nullptr;
    while(cur || !st.empty())
    {
        //左树
        while(cur)
        {
            st.push(cur);
            cur=cur->left;
        }

        TreeNode* top=st.top();
        //判断出栈。和存数据
        if(top->right==nullptr || top->right==prev)    //prev上一次pop的对象
        {
            st.pop();
            prev=top;
            v.push_back(top->val);
        }
        else    //右子树不为空或者未访问
        {
            //访问右子树
            cur=top->right;
        }
    }

    return v;
    }
};