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Binary Tree Level Order Traversal II 二叉树层序遍历之二
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]]
从底部层序遍历其实还是从顶部开始遍历，只不过最后存储的方式有所改变，可以参见我之前的博文 http://www.cnblogs.com/grandyang/p/4051321.html。 代码如下：
/** * Definition for binary tree * 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) { 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; }};
Binary Tree Level Order Traversal II 二叉树层序遍历之二
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