LeetCode144. 二叉树的前序遍历🌟🌟🌟🌟🌟简单
课后作业
问题描述
原文链接：144. 二叉树的前序遍历
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
代码实现
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) {
if(root == null){
return new ArrayList<>();
}
Stack<TreeNode> stack = new Stack();
List<Integer> res = new ArrayList();
stack.push(root);
while(!stack.isEmpty()){
TreeNode temp = stack.pop();
res.add(temp.val);
if(temp.right != null){
stack.push(temp.right);
}
if(temp.left != null){
stack.push(temp.left);
}
}
return res;
}
}
Python
class Solution(object):
def preorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
if not root:
return []
stack = []
res = []
stack.append(root)
while len(stack) > 0:
temp = stack.pop()
res.append(temp.val)
if temp.right:
stack.append(temp.right)
if temp.left:
stack.append(temp.left)
return res
C++
class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
if(!root){
return {};
}
stack<TreeNode*> st;
vector<int> res;
st.push(root);
while(!st.empty()){
TreeNode* temp = st.top();
st.pop();
res.push_back(temp->val);
if(temp->right){
st.push(temp->right);
}
if(temp->left){
st.push(temp->left);
}
}
return res;
}
};
Go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func preorderTraversal(root *TreeNode) []int {
if root == nil{
return []int{}
}
var stack []*TreeNode
var res []int
stack = append(stack, root)
for len(stack) > 0{
temp := stack[len(stack)-1]
stack = stack[:len(stack)-1]
res = append(res, temp.Val)
if temp.Right != nil{
stack = append(stack, temp.Right)
}
if temp.Left != nil{
stack = append(stack, temp.Left)
}
}
return res
}