# 问题 原题链接：[https://leetcode.com/problems/find-leaves-of-binary-tree/](https://leetcode.com/problems/find-leaves-of-binary-tree/) Given a binary tree, collect a tree's nodes as if you were doing this: Collect and remove all leaves, repeat until the tree is empty. Example: ``` Input: [1,2,3,4,5] 1 / \ 2 3 / \ 4 5 Output: [[4,5,3],[2],[1]] ``` Explanation: ``` 1. Removing the leaves [4,5,3] would result in this tree: 1 / 2 2. Now removing the leaf [2] would result in this tree: 1 3. Now removing the leaf [1] would result in the empty tree: [] ``` # 分析过程 - 输入：一棵二叉树 - 输出：每次移除所有子节点，直到整棵树都是空的 - 思路：递归判断节点是否是叶节点，对根节点就是叶节点的情况单独处理，详细见注释。 # 解决方法 ```cpp class Solution { public: void _findLeaves(TreeNode *super, TreeNode *node, int level, vector<vector<int>>& vec) { if (!node) { return; } // 判断是否是叶节点 if (!node->left && !node->right) { // 是叶节点，分别看是父节点的左还是右节点，分别处理 if (super->left == node) { super->left = nullptr; } else if (super->right == node) { super->right = nullptr; } vec[level].push_back(node->val); } else { // 不是叶节点的话，继续递归 _findLeaves(node, node->left, level, vec); _findLeaves(node, node->right, level, vec); } } vector<vector<int>> findLeaves(TreeNode* root) { vector<vector<int>> vec; if (!root) { return vec; } int level = 0; while (root->left || root->right) { vec.push_back({}); // 分别对左右子树进行剥离 _findLeaves(root, root->left, level, vec); _findLeaves(root, root->right, level, vec); level++; } vec.push_back({ root->val }); return vec; } }; ```