1080. Insufficient Nodes in Root to Leaf Paths

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1080. Insufficient Nodes in Root to Leaf Paths

Description

Given the root of a binary tree and an integer limit, delete all insufficient nodes in the tree simultaneously, and return the root of the resulting binary tree.

A node is insufficient if every root to leaf path intersecting this node has a sum strictly less than limit.

A leaf is a node with no children.

Example 1:

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Input: root = [1,2,3,4,-99,-99,7,8,9,-99,-99,12,13,-99,14], limit = 1
Output: [1,2,3,4,null,null,7,8,9,null,14]

Example 2:

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Input: root = [5,4,8,11,null,17,4,7,1,null,null,5,3], limit = 22
Output: [5,4,8,11,null,17,4,7,null,null,null,5]

Example 3:

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Input: root = [1,2,-3,-5,null,4,null], limit = -1
Output: [1,null,-3,4]

Constraints:

  • The number of nodes in the tree is in the range [1, 5000].
  • -10^5 <= Node.val <= 10^5
  • -10^9 <= limit <= 10^9

Hints/Notes

  • binary tree

Solution

Language: C++

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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:
    TreeNode* sufficientSubset(TreeNode* root, int limit) {
        if (!root) {
            return root;
        }

        if (!root->left && !root->right) {
            if (root->val < limit) {
                return nullptr;
            }
            return root;
        }

        root->left = sufficientSubset(root->left, limit - root->val);
        root->right = sufficientSubset(root->right, limit - root->val);

        if (!root->left && !root->right) {
            return nullptr;
        }
        return root;
    }
};