题目链接
题目描述
给你二叉树的根节点 root
,返回它节点值的 前序遍历。
示例 1:
输入: root = [1,null,2,3]
输出: [1,2,3]
示例 2:
输入: root = []
输出: []
示例 3:
输入: root = [1]
输出: [1]
示例 4:
输入: root = [1,2]
输出: [1,2]
示例 5:
输入: root = [1,null,2]
输出: [1,2]
提示:
- 树中节点数目在范围
[0, 100]
内 -100 <= Node.val <= 100
解题思路
方法一:递归
/**
* 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> preorderTraversal(TreeNode* root) {
vector<int> ans;
preorder(root,ans);
return ans;
}
void preorder(TreeNode* root,vector<int>& ans){
if(root==nullptr){
return;
}
ans.push_back(root->val);
preorder(root->left,ans);
preorder(root->right,ans);
}
};
- 时间复杂度 O(n)
- 空间复杂度 O(n)
方法二:迭代
```cpp /**- 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
preorderTraversal(TreeNode root) {
}vector<int> ans; stack<TreeNode*> stk; while(!stk.empty() || root != nullptr){ while(root!=nullptr){ ans.push_back(root->val); stk.push(root); root = root->left; } root = stk.top(); stk.pop(); root = root->right; } return ans;
};
```java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
Deque<TreeNode> stack = new LinkedList<>();
List<Integer> res = new LinkedList<>();
while(root != null || !stack.isEmpty()){
if(root != null){
res.add(root.val);
stack.push(root);
root = root.left;
}else{
root = stack.poll().right;
}
}
return res;
}
}
- 时间复杂度 O(n)
-
方法三:morris遍历
/** * 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> preorderTraversal(TreeNode* root) { vector<int> ans; TreeNode* p1 = root,*p2 = nullptr; while(p1!=nullptr){ p2 = p1->left;//查看p1的左子树 if(p2!=nullptr){// p1的左子树不为空,遍历p1和左子树 while(p2->right!=nullptr&&p1!=p2->right){ p2 = p2->right; } if(p2->right==nullptr){// p1的左子树未遍历完,遍历p1和左子树 ans.push_back(p1->val); p2->right = p1; p1 = p1->left; continue; }else{// p1的左子树全部遍历完毕 p2->right = nullptr;//断开和p1连接 p1 = p1->right;//开始遍历p1左子树 } }else{// p1的左子树为空,遍历p1和右子树 ans.push_back(p1->val); p1 = p1->right; } } return ans; } };
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public List<Integer> preorderTraversal(TreeNode root) { List<Integer> res = new ArrayList<>(); if(root == null){ return res; } while(root != null){ if(root.left != null){// 左子树不为空 TreeNode tmp = root.left; while(tmp.right != null && tmp.right != root){ tmp = tmp.right; } // 将左子树最后一个节点和当前根节点相连 if(tmp.right == null){ res.add(root.val); tmp.right = root; root = root.left; }else{// 如果已经相连,说明当前根节点和左子树已经遍历完毕,断开连接并遍历右子树 tmp.right = null; root = root.right; } }else{// 左子树为空 res.add(root.val); root = root.right; } } return res; } }
时间复杂度 O(n)
- 空间复杂度 O(1)