leetcode刷題筆記(Tree相關)

104. Maximum Depth of Binary Tree

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Note: A leaf is a node with no children.

Example:

Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its depth = 3.

思路:此題用遞歸來做,找到左右子樹最深的depth,然後+1,最後做遞歸。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        if root is None: h=0
        else:
            h= 1+max(self.maxDepth(root.left), self.maxDepth(root.right))
        return h

 

 

108. Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

思路:二叉樹,左邊比中間結點小,右邊比中間節點大,那麼每次找到中心節點就是父節點,然後左右分別進行遞歸。 

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
        if not nums : return None
        mid = len(nums)//2
        root = TreeNode(nums[mid])
        root.left = self.sortedArrayToBST(nums[:mid])
        root.right = self.sortedArrayToBST(nums[mid+1:])
        return root

 

199. Binary Tree Right Side View

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

Example:

Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:

   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---

思路:還是可以用遞歸來實現,如果右邊比左邊長就全部輸出右邊,左邊比右邊長就右邊+左邊比右邊多的長度。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def rightSideView(self, root: TreeNode) -> List[int]:
        if root is None:
            return []
        else:
            lists = [root.val]
            left = lists+self.rightSideView(root.left)
            right = lists+self.rightSideView(root.right)
            if len(right)>len(left):
                return right
        return right+left[len(right):]
  

 

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