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:

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:

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:

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].
  • -105 <= Node.val <= 105
  • -109 <= limit <= 109

Solutions

Solution 1

Python Code
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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 sufficientSubset(
        self, root: Optional[TreeNode], limit: int
    ) -> Optional[TreeNode]:
        if root is None:
            return None
        limit -= root.val
        if root.left is None and root.right is None:
            return None if limit > 0 else root
        root.left = self.sufficientSubset(root.left, limit)
        root.right = self.sufficientSubset(root.right, limit)
        return None if root.left is None and root.right is None else root

Java Code
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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 {
    public TreeNode sufficientSubset(TreeNode root, int limit) {
        if (root == null) {
            return null;
        }
        limit -= root.val;
        if (root.left == null && root.right == null) {
            return limit > 0 ? null : root;
        }
        root.left = sufficientSubset(root.left, limit);
        root.right = sufficientSubset(root.right, limit);
        return root.left == null && root.right == null ? null : root;
    }
}

C++ Code
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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 nullptr;
        }
        limit -= root->val;
        if (!root->left && !root->right) {
            return limit > 0 ? nullptr : root;
        }
        root->left = sufficientSubset(root->left, limit);
        root->right = sufficientSubset(root->right, limit);
        return !root->left && !root->right ? nullptr : root;
    }
};

Go Code
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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func sufficientSubset(root *TreeNode, limit int) *TreeNode {
	if root == nil {
		return nil
	}

	limit -= root.Val
	if root.Left == nil && root.Right == nil {
		if limit > 0 {
			return nil
		}
		return root
	}

	root.Left = sufficientSubset(root.Left, limit)
	root.Right = sufficientSubset(root.Right, limit)

	if root.Left == nil && root.Right == nil {
		return nil
	}
	return root
}

TypeScript Code
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/**
 * 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 sufficientSubset(root: TreeNode | null, limit: number): TreeNode | null {
    if (root === null) {
        return null;
    }
    limit -= root.val;
    if (root.left === null && root.right === null) {
        return limit > 0 ? null : root;
    }
    root.left = sufficientSubset(root.left, limit);
    root.right = sufficientSubset(root.right, limit);
    return root.left === null && root.right === null ? null : root;
}

JavaScript Code
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/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} limit
 * @return {TreeNode}
 */
var sufficientSubset = function (root, limit) {
    if (root === null) {
        return null;
    }
    limit -= root.val;
    if (root.left === null && root.right === null) {
        return limit > 0 ? null : root;
    }
    root.left = sufficientSubset(root.left, limit);
    root.right = sufficientSubset(root.right, limit);
    return root.left === null && root.right === null ? null : root;
};