[LeetCode] 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 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).

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.

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

删除给定值的叶子节点。

给你一棵以 root 为根的二叉树和一个整数 target ，请你删除所有值为 target 的 叶子节点 。

注意，一旦删除值为 target 的叶子节点，它的父节点就可能变成叶子节点；如果新叶子节点的值恰好也是 target ，那么这个节点也应该被删除。

也就是说，你需要重复此过程直到不能继续删除。

思路一 - 递归

Java实现

/**
 * 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 removeLeafNodes(TreeNode root, int target) {
		// corner case
		if (root == null) {
			return null;
		}
		root.left = removeLeafNodes(root.left, target);
		root.right = removeLeafNodes(root.right, target);
		if (root.left == null && root.right == null && root.val == target) {
			return null;
		}
		return root;
	}
}

复杂度

时间O(n)
空间O(n)

思路二 - 后序遍历

Java实现

/**
 * 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 removeLeafNodes(TreeNode root, int target) {
        // corner case
        if (root == null) {
            return root;
        }
        return helper(root, target);
    }

    private TreeNode helper(TreeNode root, int target) {
        if (root == null) {
            return null;
        }
        TreeNode left = helper(root.left, target);
        TreeNode right = helper(root.right, target);
        root.left = left;
        root.right = right;
        if (left == null && right == null && root.val == target) {
            return null;
        }
        return root;
    }
}

复杂度

时间O(n)
空间O(n)

相关题目

814. Binary Tree Pruning
1325. Delete Leaves With a Given Value