Question

link

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],
]

Stats

Frequency 1
Difficulty 3
Adjusted Difficulty 1
Time to use ----------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

Analysis

This is the same question as previous one.

Solution

There are also 2 solution: BFS and DFS.

I post BFS code below. Only 2 lines are different: ans.get() and ans.add().

Code

public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
    ArrayList<ArrayList<Integer>> ans = new ArrayList<ArrayList<Integer>>();
    if (root == null) return ans;
    Queue<TreeNode> q = new LinkedList<TreeNode>();
    q.add(root);
    while (! q.isEmpty()) {
        ans.add(0, new ArrayList<Integer>());
        int curSize = q.size();
        for (int i = 0; i < curSize; i ++) {
            TreeNode node = q.remove();
            ans.get(0).add(node.val);
            if (node.left != null) q.add(node.left);
            if (node.right != null) q.add(node.right);
        }
    }
    return ans;
}