Binary Tree Level Order Traversal II

【题目描述】

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).

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]]

【思路】

思路其实和Binary Tree Level Order Traversal是一样的，不过是最后插入的时候上一题是从尾部插入，这一道是从头部插而已。

【代码】

 

/** * 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) {        if(root==NULL) return {};        vector<vector<int>> ans;        queue<TreeNode*> q;        q.push(root);        while(!q.empty()){            vector<int> p;            int len=q.size();            for(int i=0;i<len;i++){                TreeNode* node=q.front();                p.push_back(node->val);                if(node->left!=NULL) q.push(node->left);                if(node->right!=NULL) q.push(node->right);                q.pop();            }            ans.insert(ans.begin(),p);        }        return ans;    }};