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leetcode Binary Tree Level Order Traversal I II

时间:2014-11-02 07:06:13      阅读:195      评论:0      收藏:0      [点我收藏+]

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Given a binary tree, return the level order traversal of its nodes‘ values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   /   9  20
    /     15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> result=new  ArrayList<ArrayList<Integer>>();
        if(root==null){
            return result;
        }
        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        while(!queue.isEmpty()){
            ArrayList<Integer> level=new ArrayList<Integer>();
            int size=queue.size();
            for(int i=0;i<size;i++){
                TreeNode head=queue.poll();
                level.add(head.val);
                if(head.left!=null){
                    queue.add(head.left);
                }
                if(head.right!=null){
                    queue.add(head.right);
                }
            }
            result.add(level);
        }
        return result;

    }
}

Binary Tree Level Order Traversal  II

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]
]
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
         ArrayList<ArrayList<Integer>> result=new  ArrayList<ArrayList<Integer>>();
        if(root==null){
            return result;
        }
        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        while(!queue.isEmpty()){
            ArrayList<Integer> level=new ArrayList<Integer>();
            int size=queue.size();
            for(int i=0;i<size;i++){
                TreeNode head=queue.poll();
                level.add(head.val);
                if(head.left!=null){
                    queue.add(head.left);
                }
                if(head.right!=null){
                    queue.add(head.right);
                }
            }
            result.add(0,level);
        }
        return result;
    }
}

 

leetcode Binary Tree Level Order Traversal I II

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原文地址:http://www.cnblogs.com/lilyfindjobs/p/4068330.html

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