[LeetCode 解题报告]144. Binary Tree Preorder Traversal

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

Example:

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

Output: [1,2,3]

Follow up: Recursive solution is trivial, could you do it iteratively?

考察：二叉树的前序遍历 

Method 1. Recursive

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

Method 2.Iteratively 

/**
 * 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) {
        if (root == NULL)
            return {};
        
        stack<TreeNode*> s{{root}};
        vector<int> res;
        
        while(!s.empty()) {
            TreeNode* cur = s.top();
            s.pop();
            res.push_back(cur->val);
            
            if (cur->right)
                s.push(cur->right);
            if (cur->left)
                s.push(cur->left);
        }
        return res;
    }
};

 完，