JAVA算法练习题day36

学习了二叉树递归的写法和相关题目。

45.二叉树的右视图

先看这道：110.平衡二叉树，easy

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):def isBalanced(self, root):""":type root: Optional[TreeNode]:rtype: bool"""#记录高度并返回，用于“递”的计算。返回的高度是正数。那如果当前节点不是平衡二叉树，返回一个负数作为标记（-1）。def dfs(root):if root is None:return 0leftH = dfs(root.left)if leftH == -1 :return -1#难点在这：应该在获取到rightH时候去计算高度差来判断平衡二叉树rightH = dfs(root.right)if rightH == -1 or abs(rightH-leftH) > 1 :return -1return max(leftH,rightH)+1return dfs(root) != -1

右视图的题解：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):def rightSideView(self, root):""":type root: Optional[TreeNode]:rtype: List[int]"""#先递归右子树，问题：怎么记录答案，怎么确定当前节点是否要记录到答案中去#记录当前递归深度，如果当前递归深度与当前答案长度一致，则将当前节点加入到答案中。否则不加入。ans = []def dfs(root,height):if root is None:return if height == len(ans):ans.append(root.val)dfs(root.right,height+1)dfs(root.left,height+1)dfs(root,0)return ans