binary_tree_level_order_traversal_ii.cpp

/*
107. Binary Tree Level Order Traversal II
Easy

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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]
*/

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        queue<TreeNode*> q;
        q.push(root);
        vector<vector<int>> result;
        if(root==NULL) return result;
        result.push_back({root->val});
        while(!q.empty()){
            int size=q.size();
          vector<int> t;
            for(int i=0;i<size;i++){
              TreeNode* temp = q.front();
                q.pop();
                if(temp->left!=NULL)
                t.push_back(temp->left->val);
                if(temp->right!=NULL)
                t.push_back(temp->right->val);
                if(temp->left!=NULL)
                q.push(temp->left);
                if(temp->right!=NULL)
                q.push(temp->right);
            }
            if(t.size()!=0)
            result.push_back(t);
               
        }
        reverse(result.begin(), result.end());
        return result;
    }
};