145. 二叉树的后序遍历

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> stk;
        auto pt = root;
        while(pt || stk.size())
        {
            while(pt)
            {
                stk.push(pt);
                auto tmp = pt;
                pt = pt->left;
                tmp->left = nullptr;
            }
            if(stk.size())
            {
                pt = stk.top();
                if(!pt->left && !pt->right)
                {
                    res.push_back(pt->val);
                    stk.pop();
                    pt = nullptr;
                }
                else
                {
                    auto tmp = pt;
                    pt = pt->right;
                    tmp->right = nullptr;
                }
            }
        }
        return res;
    }
};

第二次写题
递归的写法

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    vector<int> res;
    void dfs(TreeNode* root)
    {
        if(!root) return;
        dfs(root->left);
        dfs(root->right);
        res.push_back(root->val);
    }
public:
    vector<int> postorderTraversal(TreeNode* root) {
        dfs(root);
        return res;
    }
};

迭代的写法：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        stack<TreeNode*> stk;
        vector<int> res;
        while(root || stk.size())
        {
            while(root)
            {
                stk.push(root);
                TreeNode* tmp = root;
                root = root->left;
                tmp->left = nullptr;
            }
            if(stk.size())
            {
                root = stk.top();
                if(!root->left && !root->right)
                {
                    res.push_back(root->val);
                    stk.pop();
                    root = nullptr;
                }
                else if(root->right)
                {
                    TreeNode* tmp = root;
                    root = root->right;
                    tmp->right = nullptr;
                }
            }
        }
        return res;
    }
};

Morris写法

第三次写题

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        stack<TreeNode*> stk;
        vector<int> res;
        while(root|| stk.size())
        {
            while(root)
            {
                stk.push(root);
                TreeNode* tmp = root;
                root = root->left;
                tmp->left = nullptr;
            }
            root = stk.top();
            if(!root->right)
            {
                res.push_back(root->val);
                root = nullptr;
                stk.pop();
            }
            else
            {
                TreeNode* tmp = root;
                root = root->right;
                tmp->right = nullptr;
            }
        }
        return res;
    }
};

第三次写题

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        stack<TreeNode*> stk;
        vector<int> res;
        while(root || stk.size())
        {
            while(root)
            {
                stk.push(root);
                TreeNode* tmp = root;
                root = root->left;
                tmp->left = nullptr;
            }
            root = stk.top();
            if(root->right)
            {
                TreeNode* tmp = root;
                root = root->right;
                tmp->right = nullptr;
            }
            else{
                stk.pop();
                res.push_back(root->val);
                root = nullptr;
            }
        }
        return res;
    }
};