代码随想录算法训练营第十三天 | 二叉树的递归遍历 层序遍历

144. 二叉树的前序遍历

class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []
        def dfs(node):
            if node is None:
                return None
            result.append(node.val)
            dfs(node.left)
            dfs(node.right)
        dfs(root)
        return result

145. 二叉树的后序遍历

class Solution:
    def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []
        def dfs(node):
            if node is None:
                return None
            dfs(node.left)
            dfs(node.right)
            result.append(node.val)
        dfs(root)
        return result

94. 二叉树的中序遍历

class Solution:
    def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []
        def dfs(node):
            if node is None:
                return None
            dfs(node.left)
            result.append(node.val)
            dfs(node.right)
        dfs(root)
        return result

102. 二叉树的层序遍历

class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        queue = [root]
        result = []
        while queue:
            level = []
            for _ in range(len(queue)):
                cur = queue.pop(0)
                level.append(cur.val)
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            result.append(level)
        return result

107. 二叉树的层序遍历 II

class Solution:
    def levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        queue = [root]
        result = []
        while queue:
            level = []
            for _ in range(len(queue)):
                cur = queue.pop(0)
                level.append(cur.val)
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            result.append(level)
        return result[::-1]

199. 二叉树的右视图

class Solution:
    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []
        queue = [root]
        result = []
        while queue:
            level = []
            for _ in range(len(queue)):
                cur = queue.pop(0)
                level.append(cur.val)
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            result.append(level[-1])
        return result

637. 二叉树的层平均值

class Solution:
    def averageOfLevels(self, root: Optional[TreeNode]) -> List[float]:
        if not root:
            return []
        queue = [root]
        result = []
        while queue:
            level = []
            for _ in range(len(queue)):
                cur = queue.pop(0)
                level.append(cur.val)
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            result.append(sum(level)/len(level))
        return result

429. N 叉树的层序遍历

class Solution:
    def levelOrder(self, root: 'Node') -> List[List[int]]:
        if not root:
            return []
        queue = deque([root])
        result = []
        while queue:
            level = []
            for _ in range(len(queue)):
                cur = queue.popleft()
                level.append(cur.val)
                if cur.children:
                    for child in cur.children:
                        queue.append(child)
            result.append(level)
        return result

515. 在每个树行中找最大值

class Solution:
    def largestValues(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []
        queue = deque([root])
        result = []
        while queue:
            level = []
            for _ in range(len(queue)):
                cur = queue.popleft()
                level.append(cur.val)
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            result.append(max(level))
        return result

116. 填充每个节点的下一个右侧节点指针

class Solution:
    def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
        if not root:
            return None
        queue = deque([root])
        while queue:
            level = []
            for _ in range(len(queue)):
                cur = queue.popleft()
                level.append(cur)
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            for i in range(len(level) - 1):
                level[i].next = level[i + 1]
            if level:
                level[-1].next = None
        return root

104. 二叉树的最大深度

class Solution:
    def maxDepth(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0
        queue = deque([root])
        cnt = 0
        while (queue):
            for _ in range(len(queue)):
                cur = queue.popleft()
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            cnt += 1
        return cnt

111. 二叉树的最小深度

class Solution:
    def minDepth(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0
        queue = deque([root])
        cnt = 0
        while (queue):
            cnt += 1
            for _ in range(len(queue)):
                cur = queue.popleft()
                if (not cur.left) and (not cur.right):
                    return cnt
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
        return cnt

唯一需要注意的是统计层数时进入下一层while(queue)后就要计数加一