leetcode刷题笔记1 二叉树part04

leetcode112 路径总和

class Solution:
    def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
        if not root:
            return False
        return self.traversal(root, targetSum - root.val)
    def traversal(self, cur: TreeNode, count: int):
        if not cur.left and not cur.right and count == 0:
            return True 
        if not cur.left and not cur.right:
            return False
        if cur.left:
            count -= cur.left.val
            if self.traversal(cur.left, count):
                return True
            count += cur.left.val
        if cur.right:
            count -= cur.right.val
            if self.traversal(cur.right, count):
                return True
            count += cur.right.val
        return False

leetcode106 从中序与后序遍历序列构造二叉树

从本题和下面的这两道题学习如何根据中序和前序、中序和后序遍历序列构造二叉树。同时，没有中序的情况是不能构造二叉树的

class Solution:
    def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
        if not postorder:
            return None

        root_val = postorder[-1]
        root = TreeNode(root_val)

        seperator_idx = inorder.index(root_val)

        inorder_left = inorder[:seperator_idx]
        inorder_right = inorder[seperator_idx + 1:]

        postorder_left = postorder[:len(inorder_left)]
        postorder_right = postorder[len(inorder_left):len(postorder)- 1]

        root.left = self.buildTree(inorder_left, postorder_left)
        root.right = self.buildTree(inorder_right, postorder_right)
        return root

leetcode115 从前序与中序遍历序列构造二叉树

class Solution:
    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
        if not preorder:
            return None
        root_val = preorder[0]
        root = TreeNode(root_val)
        seperator_idx = inorder.index(root_val)
        inorder_left = inorder[:seperator_idx]
        inorder_right = inorder[seperator_idx + 1:]
        preorder_left = preorder[1:1 + len(inorder_left)]
        preorder_right = preorder[1 + len(inorder_left):]
        root.left = self.buildTree(preorder_left, inorder_left)
        root.right = self.buildTree(preorder_right, inorder_right)
        return root