Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree [3,9,20,null,null,15,7]
,
3 / 9 20 / 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
102. Binary Tree Level Order Traversal 的变形,只是最后输出的顺序变了,这题是从最后一层开始。
Python:
class TreeNode(object): def __init__(self, x): self.val = x self.left = None self.right = None class Solution(object): def levelOrderBottom(self, root): """ :type root: TreeNode :rtype: List[List[int]] """ if root is None: return [] result, current = [], [root] while current: next_level, vals = [], [] for node in current: vals.append(node.val) if node.left: next_level.append(node.left) if node.right: next_level.append(node.right) current = next_level result.append(vals) return result[::-1]
C++:Iteration
class Solution { public: vector<vector<int> > levelOrderBottom(TreeNode *root) { vector<vector<int> > res; if (root == NULL) return res; queue<TreeNode*> q; q.push(root); while (!q.empty()) { vector<int> oneLevel; int size = q.size(); for (int i = 0; i < size; ++i) { TreeNode *node = q.front(); q.pop(); oneLevel.push_back(node->val); if (node->left) q.push(node->left); if (node->right) q.push(node->right); } res.insert(res.begin(), oneLevel); } return res; } };
C++:Recursion
class Solution { public: vector<vector<int>> levelOrderBottom(TreeNode* root) { vector<vector<int> > res; levelorder(root, 0, res); return vector<vector<int> > (res.rbegin(), res.rend()); } void levelorder(TreeNode *root, int level, vector<vector<int> > &res) { if (!root) return; if (res.size() == level) res.push_back({}); res[level].push_back(root->val); if (root->left) levelorder(root->left, level + 1, res); if (root->right) levelorder(root->right, level + 1, res); } };
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[LeetCode] 102. Binary Tree Level Order Traversal 二叉树层序遍历
[LeetCode] 199. Binary Tree Right Side View 二叉树的右侧视图