题目链接 : https://leetcode-cn.com/problems/binary-tree-level-order-traversal/
题目描述:
给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。
例如:
给定二叉树: [3,9,20,null,null,15,7],
3
/
9 20
/
15 7
返回其层次遍历结果：
[
[3],
[9,20],
[15,7]
]
思路:
思路1 : 迭代 ,这是典型的BFS
思路2 : 递归,类似
直接看代码!
代码:
思路一:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
if not root: return []
res = []
cur_level = [root]
while cur_level:
tmp = []
next_level = []
for node in cur_level:
tmp.append(node.val)
if node.left:
next_level.append(node.left)
if node.right:
next_level.append(node.right)
res.append(tmp)
cur_level = next_level
return res
java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) return res;
Deque<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
List<Integer> tmp = new ArrayList<>();
int cnt = queue.size();
for (int i = 0; i < cnt; i++) {
TreeNode node = queue.poll();
// System.out.println(node.val);
tmp.add(node.val);
if (node.left != null) queue.add(node.left);
if (node.right != null) queue.add(node.right);
}
res.add(tmp);
}
return res;
}
}
思路二:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
res = []
def helper(root, depth):
if not root: return
if len(res) == depth:
res.append([])
res[depth].append(root.val)
helper(root.left, depth + 1)
helper(root.right, depth + 1)
helper(root, 0)
return res
java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
helper(res, root, 0);
return res;
}
private void helper(List<List<Integer>> res, TreeNode root, int depth) {
if (root == null) return;
if (res.size() == depth) res.add(new LinkedList<>());
res.get(depth).add(root.val);
helper(res, root.left, depth + 1);
helper(res, root.right, depth + 1);
}
}
还有二叉树的前序,中序,后序,层序遍历的递归和迭代,一起打包送个你们!嘻嘻
144. 二叉树的前序遍历
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]
思路:
递归:就是依次输出根,左,右,递归下去
迭代:使用栈来完成,我们先将根节点放入栈中,然后将其弹出,依次将该弹出的节点的右节点,和左节点,注意顺序,是右,左,为什么?因为栈是先入先出的,我们要先输出右节点,所以让它先进栈.
代码:
递归:
# 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: TreeNode) -> List[int]:
res = []
def helper(root):
if not root:
return
res.append(root.val)
helper(root.left)
helper(root.right)
helper(root)
return res
迭代:
# 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: TreeNode) -> List[int]:
res = []
if not root:
return res
stack = [root]
while stack:
node = stack.pop()
res.append(node.val)
if node.right:
stack.append(node.right)
if node.left:
stack.append(node.left)
return res
145. 二叉树的后序遍历
给定一个二叉树，返回它的 后序 遍历。
示例:
输入: [1,null,2,3]
1
2
/
3
输出: [3,2,1]
思路:
递归:同理,顺序:左,右,根
迭代:这就很上面的先序一样,我们可以改变入栈的顺序,刚才先序是从右到左,我们这次从左到右,最后得到的结果取逆.
代码:
递归:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
res = []
def helper(root):
if not root:
return
helper(root.left)
helper(root.right)
res.append(root.val)
helper(root)
return res
迭代:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
res = []
if not root:
return res
stack = [root]
while stack:
node = stack.pop()
if node.left :
stack.append(node.left)
if node.right:
stack.append(node.right)
res.append(node.val)
return res[::-1]
94. 二叉树的中序遍历
给定一个二叉树，返回它的中序 遍历。
示例:
输入: [1,null,2,3]
1
2
/
3
输出: [1,3,2]
思路:
递归:顺序,左右根
非递归:这次我们用一个指针模拟过程
代码:
递归:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def inorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
res = []
def helper(root):
if not root:
return
helper(root.left)
res.append(root.val)
helper(root.right)
helper(root)
return res
迭代:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def inorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
res = []
if not root:
return res
stack = []
cur = root
while stack or cur:
while cur:
stack.append(cur)
cur = cur.left
cur = stack.pop()
res.append(cur.val)
cur = cur.right
return res
102. 二叉树的层次遍历
给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。
例如:
给定二叉树: [3,9,20,null,null,15,7],
3
/
9 20
/
15 7
返回其层次遍历结果：
[
[3],
[9,20],
[15,7]
]
思路:
非常典型的BFS
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 levelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
if not root:
return []
res,cur_level = [],[root]
while cur_level:
temp = []
next_level = []
for i in cur_level:
temp.append(i.val)
if i.left:
next_level.append(i.left)
if i.right:
next_level.append(i.right)
res.append(temp)
cur_level = next_level
return res