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;
     }
 }