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1973. Count Nodes Equal to Sum of Descendants

Description

Given the root of a binary tree, return the number of nodes where the value of the node is equal to the sum of the values of its descendants.

A descendant of a node x is any node that is on the path from node x to some leaf node. The sum is considered to be 0 if the node has no descendants.

 

Example 1:

Input: root = [10,3,4,2,1]
Output: 2
Explanation:
For the node with value 10: The sum of its descendants is 3+4+2+1 = 10.
For the node with value 3: The sum of its descendants is 2+1 = 3.

Example 2:

Input: root = [2,3,null,2,null]
Output: 0
Explanation:
No node has a value that is equal to the sum of its descendants.

Example 3:

Input: root = [0]
Output: 1
For the node with value 0: The sum of its descendants is 0 since it has no descendants.

 

Constraints:

• The number of nodes in the tree is in the range [1, 105].
• 0 <= Node.val <= 105

Solutions

• Java
• C++
• Python
• Go

• /**
   * 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 equalToDescendants(TreeNode root) {
          dfs(root);
          return ans;
      }
  
      private int dfs(TreeNode root) {
          if (root == null) {
              return 0;
          }
          int l = dfs(root.left);
          int r = dfs(root.right);
          if (l + r == root.val) {
              ++ans;
          }
          return root.val + l + r;
      }
  }
  

• /**
   * 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 equalToDescendants(TreeNode* root) {
          int ans = 0;
          function<long long(TreeNode*)> dfs = [&](TreeNode* root) -> long long {
              if (!root) {
                  return 0;
              }
              auto l = dfs(root->left);
              auto r = dfs(root->right);
              ans += l + r == root->val;
              return root->val + l + r;
          };
          dfs(root);
          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 equalToDescendants(self, root: Optional[TreeNode]) -> int:
          def dfs(root):
              if root is None:
                  return 0
              l, r = dfs(root.left), dfs(root.right)
              if l + r == root.val:
                  nonlocal ans
                  ans += 1
              return root.val + l + r
  
          ans = 0
          dfs(root)
          return ans
  
  

• /**
   * Definition for a binary tree node.
   * type TreeNode struct {
   *     Val int
   *     Left *TreeNode
   *     Right *TreeNode
   * }
   */
  func equalToDescendants(root *TreeNode) (ans int) {
  	var dfs func(*TreeNode) int
  	dfs = func(root *TreeNode) int {
  		if root == nil {
  			return 0
  		}
  		l, r := dfs(root.Left), dfs(root.Right)
  		if l+r == root.Val {
  			ans++
  		}
  		return root.Val + l + r
  	}
  	dfs(root)
  	return
  }
  

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