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1973. Count Nodes Equal to Sum of Descendants πŸ”’

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

Solution 1

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# 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

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/**
 * 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;
    }
}

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/**
 * 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;
    }
};

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/**
 * 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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