C++二叉树的前序中序后序非递归实现方法详细讲解

class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> v;
        stack<TreeNode*> st;
        TreeNode*cur = root;
        while(!st.empty()||cur)
        {
            //左路节点
            while(cur)
            {
                st.push(cur);
                v.push_back(cur->val);
                cur = cur->left;
            }
            //左路节点右子树
            TreeNode* top = st.top();
            st.pop();
            cur = top->right;//转化成子问题访问右子树
        }
        return v;
    }
};

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> v;
        stack<TreeNode*> st;
        TreeNode*cur = root;
        while(cur||!st.empty())
        {
            while(cur)
            {
                //不访问
                st.push(cur);
                cur = cur->left;
            }
            TreeNode*top = st.top();
            //进行访问
            v.push_back(top->val);
            st.pop();
            cur = top->right;
        }
        return v;
    }
};

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> v;
        stack<TreeNode*> st;
        TreeNode*cur = root;
        TreeNode*prev = nullptr;
        while(cur||!st.empty())
        {
            while(cur)
            {
                st.push(cur);
                cur = cur->left;
            }
            TreeNode*top = st.top();
            //top的右子树为空，或者右子树已经访问过了（上一个访问节点时右子树的根）那么说明右子树不用访问或者访问过了，可以访问根top
            //右子树不为空，且没有访问， 则迭代子问题访问
            if(top->right==nullptr||top->right==prev)
            {
                st.pop();
                v.push_back(top->val);
                prev = top;
            }
            else
            {
                cur = top->right;
            }
        }
        return v;
    }
};