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描述
Given a binary tree root, a node X in the tree is named good if in the path from root to X there are no nodes with a value greater than X.
Return the number of good nodes in the binary tree.
Example 1:
Input: root = [3,1,4,3,null,1,5]
Output: 4
Explanation: Nodes in blue are good.
Root Node (3) is always a good node.
Node 4 -> (3,4) is the maximum value in the path starting from the root.
Node 5 -> (3,4,5) is the maximum value in the path
Node 3 -> (3,1,3) is the maximum value in the path.
Example 2:
Input: root = [3,3,null,4,2]
Output: 3
Explanation: Node 2 -> (3, 3, 2) is not good, because "3" is higher than it.
Example 3:
Input: root = [1]
Output: 1
Explanation: Root is considered as good.
Note:
The number of nodes in the binary tree is in the range [1, 10^5].
Each node's value is between [-10^4, 10^4].
解析
根据题意，就是给出了一个二叉树 root ，找出这棵树中间有多少个节点是 good 的，题目中对“ good 节点”的定义为该节点到根结点的路径上的节点的值都小于等于该节点。思路比较简单，直接用 DFS ，在递归的时候传入当前路径中的最大值，然后和当前值进行比较，如果符合题意，则结果 result 加一，递归结束的 result 即为答案。
解答
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 __init__(self):
self.result = 0
def goodNodes(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root:
return 0
def dfs(root, maxValue):
if not root:
return
maxValue = max(maxValue, root.val)
if root.val >= maxValue:
self.result += 1
dfs(root.left, maxValue)
dfs(root.right, maxValue)
dfs(root, root.val)
return self.result
运行结果
Runtime: 300 ms, faster than 53.08% of Python online submissions for Count Good Nodes in Binary Tree.
Memory Usage: 49.5 MB, less than 9.80% of Python online submissions for Count Good Nodes in Binary Tree.
解析
不使用上面的全局变量 self.result ，直接用递归进行计算结果。
解答
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 goodNodes(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root or (not root.left and not root.right):
return 1
def dfs(root, maxValue):
if not root:
return 0
maxValue = max(maxValue, root.val)
count = 1 if root.val >= maxValue else 0
return count + dfs(root.left, maxValue) + dfs(root.right, maxValue)
return dfs(root, root.val)
运行结果
Runtime: 308 ms, faster than 41.65% of Python online submissions for Count Good Nodes in Binary Tree.
Memory Usage: 49.4 MB, less than 42.93% of Python online submissions for Count Good Nodes in Binary Tree.
原题链接：leetcode.com/problems/co…
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