Leetcode 144. Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree [1,null,2,3],

   1
    \
     2
    /
   3

return [1,2,3].

遞歸

/**
 * 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> preorderTraversal(TreeNode* root) {
        vector<int> ret;
        if (root != nullptr)
            reSolver(root, ret);
        return ret;
    }
private:
    void reSolver(TreeNode* root, vector<int>& ret) {
        ret.emplace_back(root->val);
        if (root->left != nullptr) {
            reSolver(root->left, ret);
        }
        if (root->right != nullptr) {
            reSolver(root->right, ret);
        }
    }
        
};

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> preorderTraversal(TreeNode* root) {
        vector<int> ret;
        
        stack<TreeNode*> s;
        TreeNode* p = root;
        while (p != nullptr || !s.empty()) {
            
            if (p != nullptr) {
                ret.emplace_back(p->val);
                s.push(p);
                p = p->left;
            } else {
                TreeNode* r = s.top();
                s.pop();
                p = r->right;
                
            }
        }
        
        return ret;
    }
};