【leetcode】1448. Count Good Nodes in Binary Tree

标签：str   treenode   binary   upload   ber   pre   traints   inpu   name   

题目如下：

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.

Constraints:

• 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].

解题思路：遍历树，记录当前路径出现过的最大值，与到达的节点比对即可。

代码如下：

# 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 goodNodes(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        dic = {}
        def recursive(node,num,max_val):
            if max_val <= node.val:
                max_val = node.val
                dic[num] = 1
            if node.left != None:
                recursive(node.left,num*2,max_val)
            if node.right != None:
                recursive(node.right,num*2+1,max_val)

        dic[1] = 1

        recursive(root,1,root.val)

        return len(dic)

【leetcode】1448. Count Good Nodes in Binary Tree

标签：str   treenode   binary   upload   ber   pre   traints   inpu   name   

原文地址：https://www.cnblogs.com/seyjs/p/13041708.html