Binary Tree Level Order Traversal II 解答

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,#,#,15,7},

    3
   / 
  9  20
    /  
   15   7

return its bottom-up level order traversal as:

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

Solution

Same method as "Binary Tree Level Order Traversal". Note that for ArrayList, it has function add(int index, E element).

/**
 * 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>> result = new ArrayList<List<Integer>>();
        if (root == null)
            return result;
        Deque<TreeNode> prev = new ArrayDeque<TreeNode>();
        Deque<TreeNode> current;
        TreeNode tmpNode;
        prev.addLast(root);
        while (prev.size() > 0) {
            current = new ArrayDeque<TreeNode>();
            List<Integer> tmpList = new ArrayList<Integer>();
            while (prev.size() > 0) {
                tmpNode = prev.pop();
                if (tmpNode.left != null)
                    current.addLast(tmpNode.left);
                if (tmpNode.right != null)
                    current.addLast(tmpNode.right);
                tmpList.add(tmpNode.val);
            }
            prev = current;
            result.add(0, tmpList);
        }
        return result;        
    }
}