1、题目描述
给定一个二叉树,返回它的 前序 遍历。
示例:
输入: [1,null,2,3]
1
2
/
3
输出: [1,2,3]
进阶: 递归算法很简单,你可以通过迭代算法完成吗?
2、我的代码
/**
-
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 preorderTraversal(TreeNode root) {
TreeNode* cur_node;
if(root == NULL)
return result_vec;result_vec.push_back(root->val); preorderTraversal(root->left); preorderTraversal(root->right); return result_vec;
}
vector result_vec;
};
使用stack
/**
-
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 preorderTraversal(TreeNode root) {
TreeNode* cur_node;
stack<TreeNode*> stack_node;
TreeNode* pre_node;if(root == NULL) return result_vec; cur_node = root; while(cur_node != NULL || !stack_node.empty()){ if(cur_node == NULL){ cur_node = stack_node.top(); stack_node.pop(); } result_vec.push_back(cur_node->val); if(cur_node->right != NULL) stack_node.push(cur_node->right); cur_node = cur_node->left; }; return result_vec;
}
vector result_vec;
};
3、网上较好的解法
莫里斯算法逻辑并不清晰,但可能省存储空间
4、自己可以改进的地方
a、利用stack的方法,执行效率并不高,逻辑设计的不对
b、
5、优化代码至简无可简(效率击败:63%,内存击败:100%)
/** -
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 preorderTraversal(TreeNode root) {
TreeNode* cur_node;
stack<TreeNode*> stack_node;
TreeNode* pre_node;if(root == NULL) return result_vec; cur_node = root; stack_node.push(cur_node); while(!stack_node.empty()){ cur_node = stack_node.top(); stack_node.pop(); result_vec.push_back(cur_node->val); if(cur_node->right != NULL) stack_node.push(cur_node->right); if(cur_node->left != NULL) stack_node.push(cur_node->left); }; return result_vec;
}
vector result_vec;
};
6、获得的思考
遍历顺序不一样时,入stack的逻辑是不同的,否则会影响逻辑清晰度甚至是执行效率