Given a binary tree, find all leaves and then remove those leaves. Then repeat the previous steps until the tree is empty.
Example:
Given binary tree
1 / \ 2 3 / \ 4 5
Returns [4, 5, 3], [2], [1]
.
Explanation:
1. Remove the leaves [4, 5, 3]
from the tree
1 / 2
2. Remove the leaf [2]
from the tree
1
3. Remove the leaf [1]
from the tree
[]
Returns [4, 5, 3], [2], [1]
.
Credits:
Special thanks to @elmirap for adding this problem and creating all test cases.
这道题给了我们一个二叉树,让我们返回其每层的叶节点,就像剥洋葱一样,将这个二叉树一层一层剥掉,最后一个剥掉根节点。那么题目中提示说要用DFS来做,思路是这样的,每一个节点从左子节点和右子节点分开走可以得到两个深度,由于成为叶节点的条件是左右子节点都为空,所以我们取左右子节点中较大值加1为当前节点的深度值,知道了深度值就可以将节点值加入到结果res中的正确位置了,求深度的方法我们可以参见Maximum Depth of Binary Tree中求最大深度的方法,参见代码如下:
解法一:
class Solution { public: vector<vector<int>> findLeaves(TreeNode* root) { vector<vector<int>> res; helper(root, res); return res; } int helper(TreeNode *root, vector<vector<int>> &res) { if (!root) return -1; int depth = 1 + max(helper(root->left, res), helper(root->right, res)); if (depth >= res.size()) res.resize(depth + 1); res[depth].push_back(root->val); return depth; } };
下面这种DFS方法没有用计算深度的方法,而是使用了一层层剥离的方法,思路是遍历二叉树,找到叶节点,将其赋值为NULL,然后加入leaves数组中,这样一层层剥洋葱般的就可以得到最终结果了:
解法二:
class Solution { public: vector<vector<int>> findLeaves(TreeNode* root) { vector<vector<int>> res; while (root) { vector<int> leaves; root = remove(root, leaves); res.push_back(leaves); } return res; } TreeNode* remove(TreeNode *node, vector<int> &leaves) { if (!node) return NULL; if (!node->left && !node->right) { leaves.push_back(node->val); return NULL; } node->left = remove(node->left, leaves); node->right = remove(node->right, leaves); return node; } };
本文转自博客园Grandyang的博客,原文链接:找二叉树的叶节点[LeetCode] Find Leaves of Binary Tree ,如需转载请自行联系原博主。