Leetcode 1325. Delete Leaves With a Given Value (二叉树后序遍历经典题)

原创已于 2024-02-20 16:53:30 修改 · 624 阅读

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于 2024-02-20 16:51:43 首次发布

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文章描述了解决删除二叉树中所有值为目标值的叶子节点的问题，提供了两种方法，一种是使用后序遍历和辅助函数，另一种是递归处理。解决方案确保在删除过程中，父节点值也为目标值时也进行删除。

Delete Leaves With a Given Value
Solved
Medium
Topics
Companies
Hint
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

解法1：后序遍历！
注意：root = helper(root, target); 不能直接写成helper(root, target); 因为有可能root也被改成NULL。
参考case [1, 1, 1]，这里的root也被清掉了。

/**
 * 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) {
        root = helper(root, target); //注意不能直接写成helper(root, target); 因为有可能root也被改成NULL。参考case [1, 1, 1]
        return root;
    }
private:
    TreeNode *helper(TreeNode *root, int target) {
        if (!root) return NULL;
        TreeNode *left = helper(root->left, target);
        TreeNode *right = helper(root->right, target);
        
        if (!left && !right && root->val == target) return NULL;
        root->left = left;
        root->right = right;
        return root;
    }
};

也可以直接写成:

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