二叉树的非递归遍历（leetcode）

/**
 * 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> preorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> s;
        TreeNode* tmp = root;

        while(!s.empty() || tmp){
            while(tmp){
                res.push_back(tmp->val);
                s.push(tmp);
                tmp = tmp->left;
            }
            tmp = s.top();
            s.pop();
            tmp = tmp->right;
        }
        return res;
    }
};

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        TreeNode* tmp = root;
        vector<int> res;
        stack<TreeNode*> s;
        while(!s.empty() || tmp){
            while(tmp){
                s.push(tmp);
                tmp = tmp->left;
            }
            tmp = s.top();
            s.pop();
            res.push_back(tmp->val);
            tmp = tmp->right;
        }
        return res;
    }
};

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> s;
        TreeNode* tmp = root;
        while(tmp || !s.empty()){
            while(tmp){
                res.push_back(tmp->val);
                s.push(tmp);
                tmp = tmp->right;
            }
            tmp = s.top();
            s.pop();
            tmp = tmp->left;
        }
        reverse(res.begin(),res.end());
        return res;
    }
};

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> s;
        TreeNode* tmp = root;
        TreeNode* pre = NULL;

        while(tmp || !s.empty()){
            while(tmp){
                s.push(tmp);
                tmp = tmp->left;
            }
            tmp = s.top();

            //判断tmp的右节点是否存在或者是否访问过
            //若是存在且访问过或是右节点不存在
            if(tmp->right==NULL || pre==tmp->right){
                res.push_back(tmp->val);
                s.pop();
                pre = tmp;
                tmp = NULL;
            }
            //若是未访问过
            else{
                tmp = tmp->right;
                pre = NULL;
            }
        }
        return res;
    }
};