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1448. Count Good Nodes in Binary Tree

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:

  • 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].

Solutions

  • /**
 * 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 = 0;

    public int goodNodes(TreeNode root) {
        dfs(root, -100000);
        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);
    }
}

  • /**
 * 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) {
        int ans = 0;
        function<void(TreeNode*, int)> dfs = [&](TreeNode* root, int mx) {
            if (!root) {
                return;
            }
            if (mx <= root->val) {
                ++ans;
                mx = root->val;
            }
            dfs(root->left, mx);
            dfs(root->right, mx);
        };
        dfs(root, -1e6);
        return ans;
    }
};

  • # 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: TreeNode, mx: int):
            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, -1000000)
        return ans


  • /**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func goodNodes(root *TreeNode) (ans int) {
	var dfs func(*TreeNode, 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, -10001)
	return
}

  • /**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function goodNodes(root: TreeNode | null): number {
    let ans = 0;
    const dfs = (root: TreeNode | null, mx: number) => {
        if (!root) {
            return;
        }
        if (mx <= root.val) {
            ++ans;
            mx = root.val;
        }
        dfs(root.left, mx);
        dfs(root.right, mx);
    };
    dfs(root, -1e6);
    return ans;
}

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