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].
统计二叉树中好节点的数目。
给你一棵根为 root 的二叉树，请你返回二叉树中好节点的数目。
「好节点」X 定义为：从根到该节点 X 所经过的节点中，没有任何节点的值大于 X 的值。
思路是DFS前序遍历，helper函数里除了当前节点，我们还需要带一个参数，表示当前这个路径上遇到的最大的节点值。其他部分跟一般的前序遍历的题没有太大区别。
时间O(n)
空间O(n)
Java实现
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode() {} 8 * TreeNode(int val) { this.val = val; } 9 * TreeNode(int val, TreeNode left, TreeNode right) { 10 * this.val = val; 11 * this.left = left; 12 * this.right = right; 13 * } 14 * } 15 */ 16 class Solution { 17 public int goodNodes(TreeNode root) { 18 return helper(root, root.val); 19 } 20 21 private int helper(TreeNode root, int max) { 22 if (root == null) { 23 return 0; 24 } 25 int res = root.val >= max ? 1 : 0; 26 res += helper(root.left, Math.max(max, root.val)); 27 res += helper(root.right, Math.max(max, root.val)); 28 return res; 29 } 30 }