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Problem 1325 Delete Leaves With a Given Value

Posted on May 19, 2024 2 mins

Table of Contents

Problem Statement

Question

Given a binary tree root and an integer target, delete all the leaf nodes with value target.

Note that once you delete a leaf node with value target, if its parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you cannot).

Example 1

Input: root = [1,2,3,2,null,2,4], target = 2
Output: [1,null,3,null,4]
Explanation: Leaf nodes in green with value (target = 2) are removed (Picture in left).
After removing, new nodes become leaf nodes with value (target = 2) (Picture in center).

flowchart TD id1(input) A((1)) --> B((2)) A --> C((3)) C --> D((2)) D --> E((2)) D --> F((4)) id2(Output) G((1)) -->H((3)) H --> I((4))

Example 2

Input: root = [1,3,3,3,2], target = 3
Output: [1,3,null,null,2]

flowchart TD id1(input) A((1)) --> B((3)) A --> C((3)) C --> D((3)) D --> E((3)) D --> F((2)) id2(Output) G((1)) -->H((3)) H --> I((2))

Example 3

Input: root = [1,2,null,2,null,2], target = 2
Output: [1]
Explanation: Leaf nodes in green with value (target = 2) are removed at each step.

flowchart TD id1(input) A((1)) --> B((2)) A --> C((2)) C --> D((2)) id2(Output) E((1))

Constraints

The number of nodes in the tree is in the range [1, 3000].
1 <= Node.val, target <= 1000

Solution

/**
 * 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* removeLeafNodes(TreeNode* root, int target) {
        if (!root)
            return nullptr;

        root->left = removeLeafNodes(root->left, target);
        root->right = removeLeafNodes(root->right, target);

        if (!root->left && !root->right && root->val == target)
            return nullptr;

        return root;

    }
};

Complexity Analysis

Time complexity - O(N)
Space complexity - O(H)

Explanation

1. Intuition

This can be solved using DFS.
We will traverse the tree in post-order fashion.
If the node is leaf and has the value equal to the target, Remove it.
Otherwise, return the node.

2. Implementation

If the root is null, return null.
Recursively call the function on the left and right subtree.
By this we will reach the leaf nodes.
If the leaf node has the value equal to the target, return null.
Otherwise, return the node.