[LeetCode] 1325. Delete Leaves With a Given Value

[grandyang]

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 its parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you cannot).

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.

Constraints:

• The number of nodes in the tree is in the range [1, 3000].
• 1 <= Node.val, target <= 1000

这道题给了一个二叉树和一个目标值 target，说是让删除所有结点值为 target 的叶结点，且题目中说了新生成的叶结点若值为 target 也要删除。这很明显是一个从下到上的处理过程，也就是后序遍历，因为只有删除了掉了当前的叶结点，才有可能形成新的叶结点。后序遍历用递归的方式写比较简单，首先判空，然后对左子结点调用递归函数，并将返回值更新左子结点，然后再对右子结点调用递归函数，并将返回值更新右子结点。此时若左右子结点是满足要求的叶结点的话就已经被删除了，此时判断若左右子结点为空，则当前结点也是一个叶结点，若其结点值为 target，则返回空，否则返回 root 即可，参见代码如下：

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

Github 同步地址:

#1325

参考资料：

https://leetcode.com/problems/delete-leaves-with-a-given-value/

https://leetcode.com/problems/delete-leaves-with-a-given-value/solutions/484264/java-c-python-recursion-solution/

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