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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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