107.Binary Tree Level Order Traversal II

Tags: Easy Tree

Link: https://leetcode.com/problems/binary-tree-level-order-traversal-ii/

---

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]
]

---

Answer:

/**
 * 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:
    //迭代法，时间复杂度O(n)，空间复杂度O(1)
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> result;
        if(root == nullptr) return result;

        queue<TreeNode *> cur;
        cur.push(root);

        while(!cur.empty()){
            vector<int> tmp;
            for(int i = cur.size(); i > 0; --i){
                TreeNode *p = cur.front();
                cur.pop();
                tmp.push_back(p -> val);

                if(p -> left != nullptr) cur.push(p -> left);
                if(p -> right != nullptr) cur.push(p -> right);
            }
            result.push_back(tmp);
        }
        reverse(result.begin(),result.end()); //相比于102多了这一行

        return result;
    }
};