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;
    }
};