问题
Count Good Nodes in Binary Tree Medium
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
判断当前节点是否为好节点，只需要知道该节点所有的祖先节点，跟它的兄弟节点没有关系。因此，我们可以使用DFS深度优先搜索来遍历整棵树。
当我们判断某一个节点时，如果它的父节点是好节点，那么它只要大于等于父节点，就是好节点。如果父节点不是好节点，那么就要跟找到离它最近的好的祖先节点，然后进行比较。因为比较是单向的，我们只要在DFS的过程中记录上次找到的好节点（或者好节点的值），就可以很简单地找到最近的好的祖先节点进行比较。
参考答案
public class Solution {
int cnt;
public int goodNodes(TreeNode root) {
cnt = 0;
// search from root
DFS(root, null);
return cnt;
}
void DFS(TreeNode node, TreeNode lastGoodNode) {
// current node is good or not
boolean isGood = false;
// lastGoodNode == null means current node is root and need to be counted
// when it's not root and node.val >= lastGoodNode.val then count
if (lastGoodNode == null || node.val >= lastGoodNode.val) {
cnt++;
isGood = true;
}
if (node.left != null) {
DFS(node.left, isGood ? node : lastGoodNode);
}
if (node.right != null) {
DFS(node.right, isGood ? node : lastGoodNode);
}
}
}
拓展训练
来试试这个同样使用的DFS的问题吧！
[LeetCode][Hard] Making A Large Island 最大人工岛 | Java
或者到作者的LeetCode专栏中看看，有没有其他感兴趣的问题吧！
