題目:
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.
代碼:
思路是遞歸
# 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 not root:
return 0
return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))
題目:
Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.
You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.
Example 1:
Input: Tree 1 Tree 2 1 2 / \ / \ 3 2 1 3 / \ \ 5 4 7 Output: Merged tree: 3 / \ 4 5 / \ \ 5 4 7
Note: The merging process must start from the root nodes of both trees.
代碼:
遞歸
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = Noneclass Solution:
def mergeTrees(self, t1: TreeNode, t2: TreeNode) -> TreeNode:
if t1 is None and t2 is None:
return None
elif t1 is None:
return t2
elif t2 is None:
return t1
else:
t1.val += t2.val
t1.left = self.mergeTrees(t1.left, t2.left)
t1.right = self.mergeTrees(t1.right, t2.right)
return t1