LeetCode #145. 二叉树的后序遍历 题解 C/C++

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){
     }
};

//递归
class Solution1 {
     
public:
	void postorder(TreeNode *root,vector<int>& ret) {
     
		if(!root)return ;
		postorder(root->left,ret);
		postorder(root->right,ret);
		ret.emplace_back(root->val);
	}
	vector<int> postorderTraversal(TreeNode *root) {
     
		vector<int> vec;
		if(root==nullptr)return vec;
		postorder(root,vec);
		return vec;

	}
};

//迭代
class Solution {
     
public:
	vector<int> postorderTraversal(TreeNode *root) {
     
		vector<int> vec;
		if(!root) return vec;
		stack<TreeNode *>stk;//显式维护一个栈
		TreeNode *node = root;
		TreeNode *prev = nullptr;//标记右节点已被访问
		while(node || !stk.empty()) {
     
			while(node!=nullptr) {
     
				stk.emplace(node);
				node = node->left;//先将左子树入栈
			}
			node = stk.top();
			stk.pop();
			if(node->right && node->right!=prev) {
     
				stk.emplace(node);
				node = node->right;
			}else {
     
				vec.emplace_back(node->val);
				prev = node;//避免重复访问右子树
				node = nullptr;//避免重复访问左子树
			}

		}
		return vec;
	}
};