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题目描述:(链接)
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,#,#,15,7},
3
/ 9 20
/ 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
解题思路：
递归版：
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int>> levelOrderBottom(TreeNode* root) { travel(root, 1); reverse(result.begin(), result.end()); return result; } void travel(TreeNode *root, int level) { if (!root) return; if (level > result.size()) { result.push_back(vector<int>()); } result[level - 1].push_back(root->val); travel(root->left, level + 1); travel(root->right, level + 1); } private: vector<vector<int>> result; };
迭代版：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int>> levelOrderBottom(TreeNode* root) { 13 vector<vector<int>> result; 14 if (!root) return result; 15 16 queue<TreeNode *> current, next; 17 vector<int> level; 18 current.push(root); 19 20 while (!current.empty()) { 21 while (!current.empty()) { 22 TreeNode *tmp = current.front(); 23 current.pop(); 24 level.push_back(tmp->val); 25 26 if (tmp->left != nullptr) next.push(tmp->left); 27 if (tmp->right != nullptr) next.push(tmp->right); 28 } 29 30 result.push_back(level); 31 level.clear(); 32 swap(current, next); 33 } 34 35 reverse(result.begin(), result.end()); 36 37 return result; 38 } 39 };
[LeetCode]Binary Tree Level Order Traversal II
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原文地址：http://www.cnblogs.com/skycore/p/5004734.html
