【LeetCode】C++ ：中等題 - 樹 145. 二叉樹的後序周遊

145. 二叉樹的後序周遊

難度中等515

給定一個二叉樹，傳回它的 後序 周遊。

示例:

輸入: [1,null,2,3]  
   1
    \
     2
    /
   3 

輸出: [3,2,1]      

進階: 遞歸算法很簡單，你可以通過疊代算法完成嗎？

1、疊代

/**
 * 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*> stk;
        TreeNode* prev = nullptr;
        while(root != nullptr || !stk.empty()){
            while(root != nullptr){
                stk.push(root);
                root = root->left;
            }
            root = stk.top();
            stk.pop();
            if(root->right == nullptr || root->right == prev){  //沒有右節點
                res.push_back(root->val);
                prev = root;
                root = nullptr;
            } else{
                stk.push(root);
                root = root->right;
            }
        }     
        return res;
    }
};
           

2、遞歸

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