Leetcode 1448. Count Good Nodes in Binary Tree (二叉树遍历题)

1. 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].

解法1：遍历。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int goodNodes(TreeNode* root) {
        if (!root) return 0;
        helper(root, root->val);
        return count;
    }
private:
    int count = 0;
    void helper(TreeNode *root, int maxSoFar) {
        if (!root) return;
        if (root->val >= maxSoFar) {
            count++;
            maxSoFar = root->val;
            //return;
        }
        helper(root->left, maxSoFar);
        helper(root->right, maxSoFar);
    }
};