C/C++描述 LeetCode 145. 二叉樹的後序周遊

 

  大家好，我叫亓官劼（qí guān jié ）

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]      

題解

/**
 * 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> ans;
    void recursionFun(TreeNode* root){
        if(!root)
            return;
        recursionFun(root->left);
        recursionFun(root->right);
        ans.push_back(root->val);
    }
    vector<int> postorderTraversal(TreeNode* root) {
        recursionFun(root);
        return ans;
    }
};      

/**
 * 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> ans;
        if (root == nullptr) {
            return ans;
        }

        stack<TreeNode *> stk;
        TreeNode *prev = nullptr;
        while (root != nullptr || !stk.empty()) {
            while (root != nullptr) {
                stk.emplace(root);
                root = root->left;
            }
            root = stk.top();
            stk.pop();
            if (root->right == nullptr || root->right == prev) {
                ans.emplace_back(root->val);
                prev = root;
                root = nullptr;
            } else {
                stk.emplace(root);
                root = root->right;
            }
        }
        return ans;
    }
};