leetcode 二叉树先、中、后序遍历，非递归法

1. 先序遍历

vector<int> preorderTraversal(TreeNode* root)
{
    vector<int> result;
    stack<TreeNode *> st;
    TreeNode *p = root;

    while (!st.empty() || p != nullptr)
    {
        if (p != nullptr)
        {
            result.push_back(p->val);
            st.push(p);
            p = p->left;
        }
        else
        {
            p = st.top(), st.pop();
            p = p->right;
        }
    }

    return result;
}

2. 中序遍历

vector<int> inorderTraversal(TreeNode* root)
{
    vector<int> result;
    stack<TreeNode *> st;
    TreeNode *p = root;

    while (!st.empty() || p != nullptr)
    {
        if (p != nullptr)
        {
            st.push(p);
            p = p->left;
        }
        else
        {
            p = st.top(), st.pop();
            result.push_back(p->val);
            p = p->right;
        }
    }
    return result;
}

3. 后续遍历

#if 1
    vector<int> postorderTraversal(TreeNode* root)
    {
        vector<int> result;
        if (root == nullptr)
        {
            return result;
        }

        stack<TreeNode *>stack;

        for (auto p = root; p != nullptr; p = p->left)
        {
            stack.push(p);
        }

        map<TreeNode*, bool> visit;

        while (!stack.empty())
        {
            auto node = stack.top();

            if (node->right != nullptr && !visit[node->right])
            {
                for (auto p = node->right; p != nullptr; p = p->left)
                {
                    stack.push(p);
                }
                continue;
            }
            stack.pop();
            visit[node] = true;
            result.push_back(node->val);
        }

        return result;
    }
#else
    vector<int> postorderTraversal(TreeNode* root)
    {
        vector<int> result;
        stack<TreeNode *> st;
        TreeNode *p = root;

        while (!st.empty() || p != nullptr)
        {
            if (p != nullptr)
            {
                st.push(p);
                result.insert(result.begin(), p->val);
                p = p->right;
            }
            else
            {
                p = st.top(), st.pop();
                p = p->left;
            }
        }

        return result;
    }
#endif

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