1325. Delete Leaves With a Given Value

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].

class Solution {
    public TreeNode removeLeafNodes(TreeNode root, int target) {
        // use a temporary node as the parent of root
        TreeNode head = new TreeNode(0);
        head.left = root;
        
        helper(root, head, target);
        return head.left;
    }
    
    // post-order with reference to root's parent for easy removal
    private void helper(TreeNode node, TreeNode parent, int target) {
        if (node == null)
            return;
        
        helper(node.left, node, target);
        helper(node.right, node, target);
        
        // post-order: if the leaf value equals target, remove it from its parent
        if (node.left == null && node.right == null && node.val == target) {
            if (parent.left == node)
                parent.left = null;
            else
                parent.right = null;
        }
    }
}

postorder，很巧妙，用parent监视子节点，省很多事情。

https://leetcode.com/problems/delete-leaves-with-a-given-value/discuss/487095/Java-post-order-traversal-beats-100-in-time

class Solution {
 public TreeNode removeLeafNodes(TreeNode root, int target) {
        if (root.left != null) root.left = removeLeafNodes(root.left, target);
        if (root.right != null) root.right = removeLeafNodes(root.right, target);
        return root.left == root.right && root.val == target ? null : root;
    }
}

https://leetcode.com/problems/delete-leaves-with-a-given-value/discuss/484264/JavaC%2B%2BPython-Recursion-Solution

还有大佬的recursion方法，屌