二叉树的后序遍历—leetcode145

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？

方法一：递归法 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        if(root==nullptr){
            return res;
        }
        core(root,res);
        return res;
    }

    void core(TreeNode* root,vector<int>& res){
        if(root==nullptr)
            return;
        core(root->left,res);
        core(root->right,res);
        res.push_back(root->val);
    }
};

方法二：迭代法 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        if(root==nullptr){
            return res;
        }
        stack<TreeNode*> vec;
        vec.push(root);
        while(!vec.empty()){
            TreeNode* top = vec.top();
            vec.pop();
            if(top->left!=nullptr)
                vec.push(top->left);
            if(top->right!=nullptr)
                vec.push(top->right);
            res.push_back(top->val);
        }
        reverse(res.begin(),res.end());
        return res;
    }
};