二叉树的三序遍历之非递归版

①二叉树的先序遍历

/*
  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) {
        std::stack<TreeNode*> st;
        std::vector<int> ans;
        TreeNode* p = root;
        while(p || !st.empty()){
            if(p){
                ans.push_back(p->val);
                st.push(p);
                p = p->left;
            }else{
                p = st.top(); st.pop();
                p = p->right;
            }
        }
        return ans;
    }
};

②二叉树的中序遍历

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        std::stack<TreeNode*> st;
        std::vector<int> ans;
        TreeNode* p = root;
        while(p || !st.empty()){
            if(p){
                st.push(p);
                p = p->left;
            }else{
                p = st.top(); st.pop();
                ans.push_back(p->val);
                p = p->right;
            }
        }
        return ans;
    }
};

③二叉树的后序遍历

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        std::stack<TreeNode*> st;
        std::vector<int> ans;
        std::unordered_map<TreeNode*,bool> vis;
        TreeNode* p = root;
        while(p || !st.empty()){
            if(p){
                st.push(p);
                // 标记右节点未访问过！
                vis[st.top()] = false;
                p = p->left;
            }else{
                // 只有左右节点均访问完，才能将其加入答案，因为（左->右->根）!
                while(!st.empty() && vis[st.top()]){
                    p = st.top(); st.pop();
                    ans.push_back(p->val);
                }
                if(st.empty()) break;
                p = st.top();
                // 访问右节点，将其标记为已访问！
                vis[p] = true;
                p = p->right;
            }
        }
        return ans;
    }
};