代码随想录训练营第III期--018--python

# 代码随想录训练营第III期--018--python

#  513.找树左下角的值
# def findBottomLeftValue(root) -> int:
#     # 使用一个dict储存深度和值，然后再sort
#     if not root: return None  
#     # if not root.left and not root.right: return root.val
#     res = [[root.val, 0]]
#     def dfs(root, count):
#         if not root: 
#             return 
#         if not root.left.left:
#             res.append([root.left.val,count])
#         if root.left:
#             dfs(root.left, count + 1)
#         if root.right:
#             dfs(root.right, count + 1)
#     dfs(root,0)
#     res.sort(key=lambda x:x[1])
#     # print(res)
#     return res[-1][0]
class Solution:
    def findBottomLeftValue(self, root: Optional[TreeNode]) -> int:
        q = deque([root])
        while q:
            node = q.popleft()
            if node.right:
                q.append(node.right)
            if node.left:
                q.append(node.left)
            ans = node.val
        return ans

# 作者：LeetCode-Solution
# 链接：https://leetcode.cn/problems/find-bottom-left-tree-value/solution/zhao-shu-zuo-xia-jiao-de-zhi-by-leetcode-weeh/
# 来源：力扣（LeetCode）
# 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
#  112. 路径总和  
def hashPathSum(root, sum:int):
    if not root:
        return False 
    if not root.left and not root.right:
        return sum == root.val 
    return hashPathSum(root.left, sum - root.val) or hashPathSum(root.right, sum - root.val)

#  113.路径总和ii
# 不会
#  106.从中序与后序遍历序列构造二叉树 
def buildTree(inorder, postorder):
    def createTree(inorder, postorder, n):
        if n == 0:
            return None 
        k = 0 
        while postorder[n - 1] != inorder[k]:
            k += 1
        node = TreeNode(inorder[k])
        node.right = createTree(inorder[k+1]:n), postorder[k:n-1], n - k - 1)
        node.left = createTree(inorder[0:k], postorder[0:k], k)
        return node 
    return createTree(inorder, postorder, len(postorder))


#  105.从前序与中序遍历序列构造二叉树
def buildTree(preorder, inorder):
    def create(preorder, inorder,n):
        if n == 0:
            return None 
        k = 0 
        while preorder[0] != inorder[k]:
            k += 1
        node = TreeNode(inorder[k])
        node.left = create(preorder[1:k+1], inorder[0:k],k)
        node.right = create(preorder[k+1:], inorder[k+1:], n - k - 1)
        return node 
    return create(preorder, inorder, len(inorder))
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