Binary Tree Postorder/Inorder/Preoder Traversal (non-recursive) -- LeetCode 145, 94

[Question]

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree [1,null,2,3],

   1
    \
     2
    /
   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

[Analysis]
The state of recursive calls can be simulated by stacks. The key is to find the branch (left or right) where the stack pop() happened. The preorder/inorder,/postorder functions are placed in differnt branch (in, left, right) respectively.

[Solution]
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        if (!root) return {};
        stack<TreeNode*> st;
        TreeNode* p=root;
        st.push(p);
        
        vector<int> res;
        while(!st.empty()) {
            if (st.top()==p && !p->left && !p->right) { 
                //-- leave node is reached and is at the top of stack.
                res.push_back(st.top()->val);     //--shared by all orders.
                st.pop();
                continue;
            }
            if (st.top()==p) {
                //res.push_back(st.top()->val);    // -- pre-order output here.
                p=p->left;
                if (p) st.push(p);
            }
            else if (st.top()->left==p) {
                res.push_back(st.top()->val);     // -- in-order output here
                p=st.top()->right;
                if (p) st.push(p);
            }
            else if (st.top()->right==p) {
                //res.push_back(st.top()->val);   //-- post-order output here
                p=st.top();
                st.pop();
            }
        }

        return res;
    }
};