Question

1325. Delete Leaves With a Given Value

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Description

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

Solutions

Solution 1

# 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 removeLeafNodes(
    self, root: Optional[TreeNode], target: int
  ) -> Optional[TreeNode]:
    if root is None:
      return None
    root.left = self.removeLeafNodes(root.left, target)
    root.right = self.removeLeafNodes(root.right, target)
    if root.left is None and root.right is None and root.val == target:
      return None
    return root

/**
 * 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) {
    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;
  }
}

/**
 * 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;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func removeLeafNodes(root *TreeNode, target int) *TreeNode {
  if root == nil {
    return nil
  }
  root.Left = removeLeafNodes(root.Left, target)
  root.Right = removeLeafNodes(root.Right, target)
  if root.Left == nil && root.Right == nil && root.Val == target {
    return nil
  }
  return root
}

/**
 * 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 removeLeafNodes(root: TreeNode | null, target: number): TreeNode | null {
  if (!root) {
    return null;
  }
  root.left = removeLeafNodes(root.left, target);
  root.right = removeLeafNodes(root.right, target);
  if (!root.left && !root.right && root.val == target) {
    return null;
  }
  return root;
}