Tree
Depth-First Search
Binary Tree
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
Python3 Java C++ Go TypeScript
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17 # 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
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28 /**
* 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 ;
}
}
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25 /**
* 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 ;
}
};
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19 /**
* 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
}
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25 /**
* 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)
* }