Formatted question description: https://leetcode.ca/all/1448.html

1448. 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].

Related Topics:
Tree, Depth-first Search

Solution 1. Pre-order traversal

We can simply use pre-order traversal.

When visiting a node p, we pass in the maximum value maxVal we’ve seen from the root thus far.

If p->val >= maxVal, we increment the answer.

Then we update maxVal = max(maxVal, root->val), and continue to visit p’s child nodes.

// OJ: https://leetcode.com/problems/count-good-nodes-in-binary-tree/

// Time: O(N)
// Space: O(H)
class Solution {
    int ans = 0;
    void dfs(TreeNode *root, int maxVal) {
        if (!root) return;
        if (root->val >= maxVal) ++ans;
        maxVal = max(maxVal, root->val);
        dfs(root->left, maxVal);
        dfs(root->right, maxVal);
    }
public:
    int goodNodes(TreeNode* root) {
        dfs(root, INT_MIN);
        return ans;
    }
};

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int goodNodes(TreeNode root) {
        if (root == null)
            return 0;
        int count = 1;
        Queue<TreeNode> nodeQueue = new LinkedList<TreeNode>();
        Queue<Integer> maxNode = new LinkedList<Integer>();
        nodeQueue.offer(root);
        maxNode.offer(root.val);
        while (!nodeQueue.isEmpty()) {
            TreeNode node = nodeQueue.poll();
            int max = maxNode.poll();
            TreeNode left = node.left, right = node.right;
            if (left != null) {
                nodeQueue.offer(left);
                if (left.val >= max)
                    count++;
                maxNode.offer(Math.max(left.val, max));
            }
            if (right != null) {
                nodeQueue.offer(right);
                if (right.val >= max)
                    count++;
                maxNode.offer(Math.max(right.val, max));
            }
        }
        return count;
    }
}

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