物件和物件引用
阿新 • • 發佈：2020-08-04
目錄
二叉樹遍歷
前序：根左右
中序：左根右
後序：左右根
深度優先
前序遍歷
144. 二叉樹的前序遍歷
給定一個二叉樹，返回它的 前序 遍歷。
示例:
輸入: [1,null,2,3]
1
\
2
/
3
輸出: [1,2,3]
# 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 = [] self.preorder(root,res) return res def preorder(self,root,res): if root: res.append(root.val) if root.left: self.preorder(root.left,res) if root.right: self.preorder(root.right,res)
#棧 #根左右，先將根節點入棧，在依次入右節點、左節點 class Solution: def preorderTraversal(self, root: TreeNode) -> List[int]: res = [] stack = [] stack.append(root) #根節點入棧 while stack: root = stack.pop() if root: res.append(root.val) stack.append(root.right) stack.append(root.left) return res
中序遍歷
94. 二叉樹的中序遍歷
給定一個二叉樹，返回它的中序 遍歷。
示例:
輸入: [1,null,2,3]
1
\
2
/
3
輸出: [1,3,2]
# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None #遞迴 class Solution: def inorderTraversal(self, root: TreeNode) -> List[int]: res = [] self.helper(root,res) return res def helper(self,root,res): if root: if root.left: self.helper(root.left,res) res.append(root.val) if root.right: self.helper(root.right,res)
#使用棧
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
res = []
stack = []
cur = root
while cur or stack:
while cur:
stack.append(cur)
cur = cur.left
cur = stack.pop()
res.append(cur.val)
cur = cur.right
return res
後序遍歷
145. 二叉樹的後序遍歷
給定一個二叉樹，返回它的 後序 遍歷。
示例:
輸入: [1,null,2,3]
1
\
2
/
3
輸出: [3,2,1]
# 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 postorder(root):
if root:
if root.left:
postorder(root.left)
if root.right:
postorder(root.right)
res.append(root.val)
postorder(root)
return res
#棧
#根左右 轉換為 根右左 逆序為 左右根
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
if not root:
return []
res = []
stack = []
stack.append(root)
while stack:
root = stack.pop()
res.append(root.val)
if root.left:
stack.append(root.left)
if root.right:
stack.append(root.right)
return res[::-1]
廣度優先
層次遍歷
102. 二叉樹的層序遍歷
給你一個二叉樹，請你返回其按 層序遍歷 得到的節點值。 （即逐層地，從左到右訪問所有節點）。
示例：
二叉樹：[3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回其層次遍歷結果：
[
[3],
[9,20],
[15,7]
]
#按照每層遍歷
# 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,queue = [],[]
queue.append([root])
res.append([root.val])
while queue:
root = queue.pop(0)
level,level_node = [],[]
for node in root:
if node.left:
level_node.append(node.left)
level.append(node.left.val)
if node.right:
level_node.append(node.right)
level.append(node.right.val)
if level_node:
queue.append(level_node)
if level:
res.append(level)
return res
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
if not root:
return []
res,queue = [],[]
queue.append([root])
level = 0
while queue:
root = queue.pop(0)
res.append([])
level_node = []
for node in root:
res[level].append(node.val)
if node.left:
level_node.append(node.left)
if node.right:
level_node.append(node.right)
if level_node:
queue.append(level_node)
level += 1
return res
前面的方法過於混亂，直接使用BFS即可
python
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
queue, res = [], []
if not root:
return res
queue.append(root)
while queue:
temp = []
length = len(queue)
while length>0:
node = queue.pop(0)
temp.append(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
length -= 1;
res.append(temp)
return res
java
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
Queue<TreeNode> queue = new LinkedList<TreeNode>();
List<List<Integer>> res = new ArrayList<List<Integer>>();
if(root != null) queue.offer(root);
while(!queue.isEmpty()){
List<Integer> temp = new ArrayList<Integer>();
int len = queue.size();
for(int i=0; i<len; i++){
TreeNode node = queue.poll();
temp.add(node.val);
if(node.left != null) queue.offer(node.left);
if(node.right != null) queue.offer(node.right);
}
res.add(temp);
}
return res;
}
}