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  • Binary Tree Level Order Traversal II

     1 public class Solution {
     2     public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
     3         // IMPORTANT: Please reset any member data you declared, as
     4         // the same Solution instance will be reused for each test case.
     5         ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
     6         if(root == null)
     7             return result;
     8         Stack<ArrayList<Integer>> stack = new Stack<ArrayList<Integer>>();
     9         stack = traversal(root);
    10         while(!stack.isEmpty())
    11             result.add(stack.pop());
    12         return result;
    13     }
    14     private Stack<ArrayList<Integer>> traversal(TreeNode root)
    15     {
    16         LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
    17         Stack<ArrayList<Integer>> result = new Stack<ArrayList<Integer>>();
    18         queue.add(root);
    19         while(!queue.isEmpty())
    20         {
    21             ArrayList<Integer> tmp = new ArrayList<Integer>();
    22             LinkedList<TreeNode> tmpqueue = new LinkedList<TreeNode>();
    23             while(!queue.isEmpty())
    24             {
    25                 TreeNode node = queue.poll();
    26                 tmp.add(node.val);
    27                 if(node.left != null)
    28                     tmpqueue.add(node.left);
    29                 if(node.right != null)
    30                     tmpqueue.add(node.right);
    31             }
    32             result.add(tmp);
    33             queue = tmpqueue;
    34         }
    35         return result;
    36     }
    37 }
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  • 原文地址:https://www.cnblogs.com/jasonC/p/3418133.html
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