Question

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:

• 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

Solution 1

# 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.
 * 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.
 * 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;
}