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Leetcode Binary Tree Level Order Traversal II

时间:2015-10-06 11:40:06      阅读:180      评论:0      收藏:0      [点我收藏+]

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]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.


解题思路:

BFS


Java code:

/**
 * 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;
        }
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        Stack<List<Integer>> s = new Stack<List<Integer>>();
        
        queue.add(root);
        while(!queue.isEmpty()){
            int size = queue.size();
            List<Integer> numberList = new ArrayList<Integer>();
            while(size > 0){
                TreeNode x = queue.remove();
                numberList.add(x.val);
                if(x.left != null) {
                    queue.add(x.left);
                }
                if(x.right != null){
                    queue.add(x.right);
                }
                size--;
            }
            s.push(numberList);
        }
         while(!s.isEmpty()){
             result.add(s.pop());
         }
         return result;
    }
}

Reference:

1. https://leetcode.com/discuss/52372/my-java-solution

 

 

 

Leetcode Binary Tree Level Order Traversal II

原文:http://www.cnblogs.com/anne-vista/p/4856882.html

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