107.Binary Tree Level Order Traversal II

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Question

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],

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  3
 / \
9  20
  /  \
 15   7

return its bottom-up level order traversal as:

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

Answer

Iterative
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/**
 * 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>> res = new ArrayList<>();
        if(root==null)
            return res;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        
        while(!queue.isEmpty()){
            List<Integer> temp = new ArrayList<>(); 
            int size = queue.size(); //must be initialize here, otherwise queue will be larger after offer(), will be changeable
            for(int i=0;i<size;i++){
                TreeNode node = queue.poll();
                temp.add(node.val);
                if(node.left!=null)
                    queue.offer(node.left);
                if(node.right!=null)
                    queue.offer(node.right);
            }
            res.add(0, temp);
        }
        return res;
    }
}

Time complexity: O(n), 3ms

Recursive
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/**
 * 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>> res = new ArrayList<>();
        if(root==null)
            return res;
        recursiveOrder(res, root, 1);
        return res;
    }
    public void recursiveOrder(List<List<Integer>> res, TreeNode node, int depth){
        if(node==null)
            return;    // to end
        if(res.size() == depth-1){
            List<Integer> temp1 = new ArrayList<>();
            temp1.add(node.val);
            res.add(0, temp1);
        }
        else if(res.size() >= depth){
            // res.get(0).add(node.val);
            res.get(res.size() - depth).add(node.val); // important get index: res.size()-depth
        }
            
        recursiveOrder(res, node.left, depth + 1);
        recursiveOrder(res, node.right, depth + 1);
    }
}

Time complexity: O(n), 1ms