Leetcode 1080. Insufficient nodes on the root to leaf path

Leetcode 1080. Insufficient nodes on the root to leaf path

topic

给定一棵二叉树的根 root，请你考虑它所有 从根到叶的路径：从根到任何叶的路径。（所谓一个叶子节点，就是一个没有子节点的节点）

假如通过节点 node 的每种可能的 “根-叶” 路径上值的总和全都小于给定的 limit，则该节点被称之为「不足节点」，需要被删除。

请你删除所有不足节点，并返回生成的二叉树的根。

Example 1:

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

输出：[1,2,3,4,null,null,7,8,9,null,14]

Example 2:

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

输出：[5,4,8,11,null,17,4,7,null,null,null,5]

Example 3:

输入：root = [5,-6,-6], limit = 0
输出：[]

Ideas

• Recursion is preferred for tree problems
• dfs uses post-order traversal, there are 2 reasons

1. Need to record the sum of the previous nodes, if you use the preorder, you have to traverse again
2. Post-order can prevent broken link

• When the left and right are insufficient nodes, the root must be the insufficient node
• The left is an insufficient node, the right does not exist, and the root is also an insufficient node
• The left does not exist, the right is an insufficient node, and the root is also an insufficient node
• Note that when root is empty, sum <limit is used to judge, and you will know the specific reason by drawing a picture.

Code

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool dfs(TreeNode*& root, int sum, int limit) {
        if (!root) return sum < limit;

        bool left = dfs(root->left, sum + root->val, limit);
        bool right = dfs(root->right, sum + root->val, limit);

        if (left && right) {
            root = NULL;
            return true;
        }

        if (left && !root->right) {
            root = NULL;
            return true;
        }

        if (!root->left && right) {
            root = NULL;
            return true;
        }

        return false;
    }

    TreeNode* sufficientSubset(TreeNode* root, int limit) {
        dfs(root, 0, limit);
        return root;
    }
};