# [LeetCode] 102. Binary Tree Level Order Traversal 二叉树层序遍历

2021年09月15日 阅读数：1

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).html

For example:
Given binary tree `{3,9,20,#,#,15,7}`,java

```    3
/ \
9  20
/  \
15   7
```

return its level order traversal as:node

```[
[3],
[9,20],
[15,7]
]```

Java: Iteration, T: O(n), S: O(1)ide

```public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List result = new ArrayList();

if (root == null) {
return result;
}

Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);

while (!queue.isEmpty()) {
ArrayList<Integer> level = new ArrayList<Integer>();
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode head = queue.poll();
if (head.left != null) {
}
if (head.right != null) {
}
}
}

return result;
}
}　　```

Java: Recursion, T: O(n), S: O(n)post

```public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
traverse(root, 1, result);
return result;
}

void traverse(TreeNode root, int level,
List<List<Integer>> result) {
if (root == null) return;

if (level > result.size())

traverse(root.left, level+1, result);
traverse(root.right, level+1, result);
}
}
```

Python: Iterationurl

```class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None

class Solution:
def levelOrder(self, root):
if root is None:
return []
result, current = [], [root]
while current:
next_level, vals = [], []
for node in current:
vals.append(node.val)
if node.left:
next_level.append(node.left)
if node.right:
next_level.append(node.right)
current = next_level
result.append(vals)
return result　　```

Python:code

```class Solution(object):
def levelTraversal(self, root):
if not root:
return []
result, queue = [], [root]
while queue:
vals = []
for i in range(len(queue)):
node = queue.pop(0)
vals.append(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
if vals:
result.append(vals)

return result　```

Python: Recursivehtm

```class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
self.right = None

class Solution(object):
def levelTraversal(self, root):
res = []
self.recur(root, 1, res)
return res

def recur(self, node, level, res):
if not node:
return

if len(res) < level:
res.append([])
res[level - 1].append(node.val)
self.recur(node.left, level + 1, res)
self.recur(node.right, level + 1, res)

if __name__ == '__main__':
node = TreeNode(3)
node.left = TreeNode(9)
node.right = TreeNode(20)
node.right.left = TreeNode(15)
node.right.right = TreeNode(7)
print Solution().levelTraversal(node)　```

Python: Recursive

```class Solution:
# @param root, a tree node
# @return a list of lists of integers
def preorder(self, root, level, res):
if root:
if len(res) < level+1: res.append([])
res[level].append(root.val)
self.preorder(root.left, level+1, res)
self.preorder(root.right, level+1, res)

def levelOrder(self, root):
res=[]
self.preorder(root, 0, res)
return res
```

C++:

```class Solution {
public:
vector<vector<int> > levelOrder(TreeNode *root) {
vector<vector<int> > res;
if (root == NULL) return res;

queue<TreeNode*> q;
q.push(root);
while (!q.empty()) {
vector<int> oneLevel;
int size = q.size();
for (int i = 0; i < size; ++i) {
TreeNode *node = q.front();
q.pop();
oneLevel.push_back(node->val);
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
res.push_back(oneLevel);
}
return res;
}
};
```

C++: Recursion

```class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int> > res;
levelorder(root, 0, res);
return res;
}
void levelorder(TreeNode *root, int level, vector<vector<int> > &res) {
if (!root) return;
if (res.size() == level) res.push_back({});
res[level].push_back(root->val);
if (root->left) levelorder(root->left, level + 1, res);
if (root->right) levelorder(root->right, level + 1, res);
}
};
```

[LeetCode] 107. Binary Tree Level Order Traversal II 二叉树层序遍历 II

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