README_EN.md

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 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
• The given binary tree will have between 1 and 3000 nodes.
• Each node's value is between [1, 1000].

Solutions

Python3

# 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

Java

/**
 * 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) {
        TreeNode p = new TreeNode(0, root, null);
        dfs(root, p, target);
        return p.left;
    }

    private void dfs(TreeNode root, TreeNode prev, int target) {
        if (root == null) {
            return;
        }
        dfs(root.left, root, target);
        dfs(root.right, root, target);
        if (root.left == null && root.right == null && root.val == target) {
            if (prev.left == root) {
                prev.left = null;
            } else {
                prev.right = null;
            }
        }
    }
}

C++

/**
 * 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) {
        TreeNode* p = new TreeNode(0, root, nullptr);
        dfs(root, p, target);
        return p->left;
    }

    void dfs(TreeNode* root, TreeNode* prev, int target) {
        if (!root) return;
        dfs(root->left, root, target);
        dfs(root->right, root, target);
        if (!root->left && !root->right && root->val == target)
        {
            if (prev->left == root) prev->left = nullptr;
            else prev->right = nullptr;
        }
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func removeLeafNodes(root *TreeNode, target int) *TreeNode {
	p := &TreeNode{0, root, nil}
	var dfs func(root, prev *TreeNode)
	dfs = func(root, prev *TreeNode) {
		if root == nil {
			return
		}
		dfs(root.Left, root)
		dfs(root.Right, root)
		if root.Left == nil && root.Right == nil && root.Val == target {
			if prev.Left == root {
				prev.Left = nil
			} else {
				prev.Right = nil
			}
		}
	}
	dfs(root, p)
	return p.Left
}

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