README_EN.md

1448. Count Good Nodes in Binary Tree

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Description

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:

<li>The number of nodes in the binary tree is in the range&nbsp;<code>[1, 10^5]</code>.</li>

<li>Each node&#39;s value is between <code>[-10^4, 10^4]</code>.</li>

Solutions

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def goodNodes(self, root: TreeNode) -> int:
        def dfs(root, mx):
            if root is None:
                return
            nonlocal ans
            if mx <= root.val:
                ans += 1
                mx = root.val
            dfs(root.left, mx)
            dfs(root.right, mx)

        ans = 0
        dfs(root, -10000)
        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 {
    private int ans;

    public int goodNodes(TreeNode root) {
        ans = 0;
        dfs(root, -10000);
        return ans;
    }

    private void dfs(TreeNode root, int mx) {
        if (root == null) {
            return;
        }
        if (mx <= root.val) {
            ++ans;
            mx = root.val;
        }
        dfs(root.left, mx);
        dfs(root.right, mx);
    }
}

C++

/**
 * 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 ans;

    int goodNodes(TreeNode* root) {
        ans = 0;
        dfs(root, -10000);
        return ans;
    }

    void dfs(TreeNode* root, int mx) {
        if (!root) return;
        if (mx <= root->val) {
            ++ans;
            mx = root->val;
        }
        dfs(root->left, mx);
        dfs(root->right, mx);
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func goodNodes(root *TreeNode) int {
	ans := 0
	var dfs func(root *TreeNode, mx int)
	dfs = func(root *TreeNode, mx int) {
		if root == nil {
			return
		}
		if mx <= root.Val {
			ans++
			mx = root.Val
		}
		dfs(root.Left, mx)
		dfs(root.Right, mx)
	}
	dfs(root, -10000)
	return ans
}

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