1325. Delete Leaves With a Given Value

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 it's parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you can't).

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).

Example 2:

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

Example 4:

Input: root = [1,1,1], target = 1
Output: []

Example 5:

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

Constraints:

• 1 <= target <= 1000

• Each tree has at most 3000 nodes.

• Each node's value is between [1, 1000].

Solutions

1. postorder traversal with recursion

/**
 * 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 (!roota) return nullptr;
        root->left = removeLeafNodes(root->left, target);
        root->right = removeLeafNodes(root->right, target);
        if (!root->left && !root->right && root->val == target)
            return nullptr;
        else
            return root;
    }
};

1. postorder traversal with stack

/**
 * 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) {
        stack<TreeNode *> s;
        TreeNode * prev = nullptr, * head = root;

        while (root || s.size()) {
            while (root) {
                s.push(root);
                root = root->left;
            }
            root = s.top();
            if (root->right && root->right != prev)
                root = root->right;
            else {
                s.pop();
                if (root->val == target && !root->left && !root->right) {
                    if (s.empty()) return nullptr;
                    else if (s.top()->left == root)
                        s.top()->left = nullptr;
                    else
                        s.top()->right = nullptr;
                }
                prev = root;
                root = nullptr;
            }
        }

        return head;
    }
};