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Formatted question description: https://leetcode.ca/all/1325.html

1325. Delete Leaves With a Given Value (Medium)

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 targetif 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].

Related Topics:
Tree

Solution 1.

  • /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public TreeNode removeLeafNodes(TreeNode root, int target) {
            Map<TreeNode, TreeNode> childParentMap = new HashMap<TreeNode, TreeNode>();
            Queue<TreeNode> queue = new LinkedList<TreeNode>();
            queue.offer(root);
            while (!queue.isEmpty()) {
                TreeNode node = queue.poll();
                TreeNode left = node.left, right = node.right;
                if (left != null) {
                    childParentMap.put(left, node);
                    queue.offer(left);
                }
                if (right != null) {
                    childParentMap.put(right, node);
                    queue.offer(right);
                }
            }
            boolean flag = true;
            while (flag) {
                flag = false;
                Set<TreeNode> keySet = childParentMap.keySet();
                Set<TreeNode> removeSet = new HashSet<TreeNode>();
                for (TreeNode node : keySet) {
                    if (isLeaf(node) && node.val == target) {
                        TreeNode parent = childParentMap.get(node);
                        if (parent == null)
                            return null;
                        else {
                            if (node == parent.left)
                                parent.left = null;
                            else if (node == parent.right)
                                parent.right = null;
                            removeSet.add(node);
                            flag = true;
                        }
                    }
                }
                for (TreeNode node : removeSet)
                    childParentMap.remove(node);
            }
            if (isLeaf(root) && root.val == target)
                root = null;
            return root;
        }
    
        public boolean isLeaf(TreeNode node) {
            return node != null && node.left == null && node.right == null;
        }
    }
    
    ############
    
    /**
     * 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;
        }
    }
    
  • // OJ: https://leetcode.com/problems/delete-leaves-with-a-given-value/
    // Time: O(N)
    // Space: O(H)
    class Solution {
    private:
        TreeNode* postorder(TreeNode *node, int target) {
            if (!node) return NULL;
            node->left = postorder(node->left, target);
            node->right = postorder(node->right, target);
            return !node->left && !node->right && node->val == target ? NULL : node;
        }
    public:
        TreeNode* removeLeafNodes(TreeNode* root, int target) {
            return postorder(root, target);
        }
    };
    
  • # 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(root, prev):
                if root is None:
                    return
                dfs(root.left, root)
                dfs(root.right, root)
                if root.left is None and root.right is None and root.val == target:
                    if prev.left == root:
                        prev.left = None
                    else:
                        prev.right = None
    
            p = TreeNode(val=0, left=root)
            dfs(root, p)
            return p.left
    
    ############
    
    # 1325. Delete Leaves With a Given Value
    # https://leetcode.com/problems/delete-leaves-with-a-given-value/
    
    # 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 dfs(self, root, target):
            if not root: return None
            
            root.left = self.dfs(root.left, target)
            root.right = self.dfs(root.right, target)
            
            if root.left == root.right == None and root.val == target: return None
            
            return root
            
                
            
            
        def removeLeafNodes(self, root: TreeNode, target: int) -> TreeNode:
            return self.dfs(root, target)
    
  • /**
     * 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;
    }
    
    

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