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 target, if 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;
}

1325 - Delete Leaves With a Given Value

1325. Delete Leaves With a Given Value (Medium)

Solution 1.

All Problems

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1325. Delete Leaves With a Given Value (Medium)

Solution 1.

All Problems

All Solutions