Approach 1: DFS

# 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]:
        def dfs(node):
            if not node:
                return None
            
            node.left = dfs(node.left)
            node.right = dfs(node.right)
 
            if node.val == target and not (node.left or node.right):
                return None
 
            return node
        
        return dfs(root)
    

Complexity

Time:
Space:

Notes

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