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？

/**
 * 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:
    void dfs(TreeNode* root, vector<int>& ans){
        if(root == NULL) return;
        if(root->left) dfs(root->left, ans);
        if(root->right) dfs(root->right, ans);
        ans.push_back(root->val);
    }
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> ans;
        dfs(root, ans);
        return ans;
    }
};

迭代的实现：用栈的方式来实现

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
#include<stack>
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        stack<TreeNode*> s;
        s.push(root);
        vector<int> ans;
        if(root == NULL) return ans;
        while(!s.empty()){
            TreeNode* temp = s.top();
            if(temp->left){
                s.push(temp->left);
                temp->left = NULL;
            } else if(temp->right){
                s.push(temp->right);
                temp->right = NULL;
            } else{
                ans.push_back(temp->val);
                s.pop();
            }
        }
        return ans;
    }
};