LeetCode 107. Binary Tree Level Order Traversal II

107. Binary Tree Level Order Traversal II

一、问题描述

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).

二、输入输出

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] ]

三、解题思路

二叉树-层次遍历

非常典型的二叉树的层次遍历。其实就是广度优先搜索 + queue 和广度优先搜索稍微不同的是，在层次遍历中需要记录下三个参数：当前层次的总个数，当前层次遍历的个数，下一层的节点个数。最初的时候，只把root入队即可。题目要求是从叶节点到根节点，所以ret是在begin的位置进行插入的。 /** * 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) { vector<vector<int>> ret; if(root == nullptr)return ret; queue<TreeNode *> q; q.push(root); int currLevelCount = 1; int nextLevelCount = 0; int count = 0; vector<int> each; while(!q.empty()) { TreeNode* head = q.front();q.pop(); count++; each.emplace_back(head->val);//加入到数组中，层次遍历中的每一行 if(head->left){ nextLevelCount++; q.push(head->left); } if(head->right){ nextLevelCount++; q.push(head->right); } if(count == currLevelCount){//当前这一层遍历完了，准备下一层的遍历 count = 0; currLevelCount = nextLevelCount; nextLevelCount = 0; ret.insert(ret.begin(),each);//题目要求从叶到根，所以插入到ret begin的位置 each.clear(); } } return ret; } };