Binary Tree Level Order Traversal II

• LeetCode: Binary Tree Level Order Traversal II

• Problem:

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).

• Examples:

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]
]

• Analysis:

Use a queue to store nodes of every level sequentially, and traverse the nodes in the queue.

• Code - Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> results = new ArrayList<List<Integer>>();
        if (root == null) {
            return results;
        }
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> temp = new ArrayList<Integer>();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                temp.add(node.val);
                if (node.left != null) {
                    queue.add(node.left);
                }
                if (node.right != null) {
                    queue.add(node.right);
                }
            }
            results.add(temp);
        }
        Collections.reverse(results);
        return results;
    }
}

• Code - C++:

/**
 * 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>> results;
        if (root == NULL) {
            return results;
        }
        queue<TreeNode*> queues;
        queues.push(root);
        while (!queues.empty()) {
            int size = queues.size();
            vector<int> temp;
            for (int i = 0; i < size; i++) {
                TreeNode* node = queues.front();
                temp.push_back(node->val);
                if (node->left != NULL) {
                    queues.push(node->left);
                }
                if (node->right != NULL) {
                    queues.push(node->right);
                }
                queues.pop();
            }
            results.push_back(temp);
        }
        reverse(results.begin(), results.end());
        return results;
    }
};

• Code - Python:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
import Queue
class Solution(object):
    def levelOrderBottom(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        results = []
        if root == None:
            return results
        queue = Queue.Queue()
        queue.put(root)
        while not queue.empty():
            size = queue.qsize()
            temp = []
            for i in range(size):
                node = queue.get()
                temp.append(node.val)
                if node.left != None:
                    queue.put(node.left)
                if node.right != None:
                    queue.put(node.right)
            results.append(temp)
        return results[::-1]

• Time Complexity: O(N), N is the number of nodes

• Space Complexity: O(M), M is the max number of nodes of a level