【LeetCode题解】144_二叉树的前序遍历
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?
方法一:递归
Java 代码
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
preorderTraversal(root, res);
return res;
}
private void preorderTraversal(TreeNode root, List<Integer> res) {
if (root == null) {
return;
}
res.add(root.val);
preorderTraversal(root.left, res);
preorderTraversal(root.right, res);
}
}
复杂度分析:
- 时间复杂度:\(O(n)\),其中,\(n\) 为二叉树节点的数目
- 空间复杂度:\(O(n)\)
Python 代码
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def preorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
def dfs(root, ret):
if root is None:
return
ret.append(root.val)
dfs(root.left, ret)
dfs(root.right, ret)
ret = list()
dfs(root, ret)
return ret
复杂度分析同上。
方法二:非递归(使用栈)
Java 代码
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
if (root == null) {
return res;
}
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode cur = stack.pop();
res.add(cur.val);
if (cur.right != null) {
stack.push(cur.right);
}
if (cur.left != null) {
stack.push(cur.left);
}
}
return res;
}
}
复杂度分析:
- 时间复杂度:\(O(n)\),其中,\(n\) 为二叉树节点的数目
- 空间复杂度:\(O(h)\),其中,\(h\) 为二叉树的高度
Python 代码
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def preorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
if root is None:
return []
ret, stack = [], [root]
while len(stack) > 0:
node = stack.pop()
ret.append(node.val)
if node.right is not None:
stack.append(node.right)
if node.left is not None:
stack.append(node.left)
return ret
复杂度分析同上。