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;    }};