LeetCode 145. 二叉树的后序遍历

给定一个二叉树，返回它的 后序 遍历。

示例:

输入: [1,null,2,3]
1

2
/
3

输出: [3,2,1]
进阶: 递归算法很简单，你可以通过迭代算法完成吗？

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-postorder-traversal

迭代：

/**
 * 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> ans;
        if(!root) return ans;
        stack<TreeNode*> nodeStack1,nodeStack2;
        TreeNode* node;
        nodeStack1.push(root);
        while( !nodeStack1.empty())
        {
            node = nodeStack1.top();
            nodeStack1.pop();
            nodeStack2.push(node);
            if(node->left) 
                nodeStack1.push(node->left);
            if(node->right)
                nodeStack1.push(node->right);
        }

        while(!nodeStack2.empty())
        {
            ans.push_back(nodeStack2.top()->val);
            nodeStack2.pop();
        }

        return ans;
    }
};

优化：来源：

执行用时为 0 ms 的范例

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> ans;
        if(root == NULL){
            return ans;
        }
        
        stack<TreeNode*> q;
        TreeNode *p = root;
        TreeNode *pLast = NULL;
        while(p){
            q.push(p);
            p = p->left;
        }
        while(!q.empty()){
            p = q.top();
            q.pop();
            // 右无或者右看过，可以看中
            if(!p->right || p->right==pLast){
                ans.push_back(p->val);
                pLast = p;
            }else{
                // 进入右
                q.push(p);
                p = p->right;
                while(p){
                    q.push(p);
                    p = p->left;
                }
            }
        }
        
        return ans;
    }
};