Question

Formatted question description: https://leetcode.ca/all/107.html

107	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,#,#,15,7},
    3
   / \
  9  20
    /  \
   15   7
return its bottom-up level order traversal as:
[
  [15,7],
  [9,20],
  [3]
]

@tag-tree

Algorithm

The traversal from the bottom sequence is actually traversal from the top, but the final storage method has changed.

Code

Java

import java.util.*;

public class Binary_Tree_Level_Order_Traversal_II {

	/**
	 * Definition for a binary tree node.
	 * public class TreeNode {
	 *     int val;
	 *     TreeNode left;
	 *     TreeNode right;
	 *     TreeNode(int x) { val = x; }
	 * }
	 */
	public class Solution {
	    public List<List<Integer>> levelOrderBottom(TreeNode root) {

	    	List<List<Integer>> list = new ArrayList<List<Integer>>();

	    	if (root == null) {
	    		return list;
	    	}

	    	Queue<TreeNode> sk = new LinkedList<>();

	    	sk.offer(root);
	    	sk.offer(null);// @note: use null as marker for end of level

	    	List<Integer> oneLevel = new ArrayList<>();
	    	while (!sk.isEmpty()) {

	    		TreeNode current = sk.poll();

	    		if (current == null) {
	    			List<Integer> copy = new ArrayList<>(oneLevel);
	    			list.add(0, copy); // diff of I and II: insert to head

	    			// clean after one level recorded
	    			oneLevel.clear();// @memorize: this api

	    			// @note:@memorize: if stack is now empty then DO NOT add null, or else infinite looping
	    			// sk.offer(null); // add marker
	    			if (!sk.isEmpty()) {
	    				sk.offer(null); // add marker
	    			}

	    			continue;
	    		}

	    		oneLevel.add(current.val);

	    		// @note:@memorize: since using null as marker, then must avoid adding null when children are null
	    		// sk.offer(current.left);
	    		// sk.offer(current.right);
	    		if (current.left != null) {
	    			sk.offer(current.left);
	    		}
	    		if (current.right != null) {
	    			sk.offer(current.right);
	    		}

	    	}

	    	return list;
	    }
	}

	public class Solution_dummyNodeAsMarker {
	    public  List<List<Integer>> levelOrderBottom(TreeNode root) {

	        LinkedList<List<Integer>> result = new LinkedList<List<Integer>>();

	        if(root == null) {
	        	return result;
			}

			// Array deques have no capacity restrictions, they grow as necessary
			Queue<TreeNode> queue = new ArrayDeque<TreeNode>(); // @note: another implementation
	        queue.add(root);
	        TreeNode dummy = new TreeNode(0);

	        queue.add(dummy);

	        List<Integer> currentLevel = new ArrayList<Integer>();
	        while(queue.size() > 1){
	        	TreeNode current = queue.poll();

	        	if (current.hashCode() != dummy.hashCode()) { // more specific with hashCode()
	        		currentLevel.add(current.val);
	        		if(current.left != null) queue.add(current.left);
	            	if(current.right != null) queue.add(current.right);
	        	} else {
	        		result.addFirst(new ArrayList<Integer>(currentLevel));
	        		currentLevel.clear();
	        		queue.add(dummy);
	        	}
	        }

	        // @note: LinkedList has API addFirst()
	        result.addFirst(currentLevel);

	        // or call reverse()
			// Collections.reverse(result);

	        return result;
	    }
	}

}

// OJ: https://leetcode.com/problems/binary-tree-level-order-traversal-ii/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        if (!root) return {};
        vector<vector<int>> ans;
        queue<TreeNode*> q;
        q.push(root);
        while (q.size()) {
            int cnt = q.size();
            ans.emplace_back();
            while (cnt--) {
                root = q.front();
                q.pop();
                ans.back().push_back(root->val);
                if (root->left) q.push(root->left);
                if (root->right) q.push(root->right);
            }
        }
        reverse(begin(ans), end(ans));
        return ans;
    }
};

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]:
        ans = []
        if root is None:
            return ans
        q = deque([root])
        while q:
            t = []
            for _ in range(len(q)):
                node = q.popleft()
                t.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            ans.append(t)
        return ans[::-1]

############

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
from collections import deque


class Solution(object):
  def levelOrderBottom(self, root):
    """
    :type root: TreeNode
    :rtype: List[List[int]]
    """
    if not root:
      return []
    ans = [[root.val]]
    queue = deque([root])
    while queue:
      levelans = []
      for _ in range(0, len(queue)):
        root = queue.popleft()
        if root.left:
          levelans.append(root.left.val)
          queue.append(root.left)
        if root.right:
          levelans.append(root.right.val)
          queue.append(root.right)
      if levelans:
        ans.append(levelans)
    return ans[::-1]

107 - Binary Tree Level Order Traversal II

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