C++ 非递归实现二叉树的前中后序遍历

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:
	//前序遍历
	vector<int> preorderTraversal(TreeNode* root) {
		stack<TreeNode*> st; //辅助栈
		vector<int> ret; //用于存放前序遍历的结果
		TreeNode* cur = root;
		while (cur || !st.empty())
		{
			//1、将左路结点入栈，入栈的同时访问左路结点
			while (cur)
			{
				st.push(cur);
				ret.push_back(cur->val);
				cur = cur->left;
			}
			//2、取出栈顶结点
			TreeNode* top = st.top();
			st.pop();
			//3、准备访问其右子树
			cur = top->right;
		}
		return ret; //返回前序遍历结果
	}
};

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:
	//中序遍历
	vector<int> inorderTraversal(TreeNode* root) {
		stack<TreeNode*> st; //辅助栈
		vector<int> ret; //用于存放中序遍历的结果
		TreeNode* cur = root;
		while (cur || !st.empty())
		{
			//1、将左路结点入栈
			while (cur)
			{
				st.push(cur);
				cur = cur->left;
			}
			//2、取出栈顶结点并访问
			TreeNode* top = st.top();
			st.pop();
			ret.push_back(top->val);
			//3、准备访问其右子树
			cur = top->right;
		}
		return ret; //返回中序遍历结果
	}
};

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:
	//后序遍历
	vector<int> postorderTraversal(TreeNode* root) {
		stack<TreeNode*> st; //辅助栈
		vector<int> ret; //用于存放后序遍历的结果
		TreeNode* cur = root;
		TreeNode* prev = nullptr; //记录上一次访问的结点
		while (cur || !st.empty())
		{
			//1、将左路结点入栈
			while (cur)
			{
				st.push(cur);
				cur = cur->left;
			}
			//2、取出栈顶结点
			TreeNode* top = st.top();
			//3、若取出结点的右子树为空，或右子树已经访问过了，则访问该结点
			if (top->right == nullptr || top->right == prev)
			{
				//访问top结点后将其从栈中弹出
				st.pop();
				ret.push_back(top->val);
				//更新上一次访问的结点为top
				prev = top;
			}
			else //4、若取出结点的右子树还未被访问，则准备访问其右子树
			{
				cur = top->right;
			}
		}
		return ret; //返回后序遍历结果
	}
};