20150624 lintcode 总结 binary-tree-level-order-traversal **-优快云博客

本文链接：https://blog.youkuaiyun.com/u014748614/article/details/46629097

本文介绍了一种使用单队列实现的二叉树层次遍历算法，通过BFS模板来解决二叉树节点值的层次遍历问题，并提供了一个具体的示例。

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Binary Tree Level Order Traversal

http://www.lintcode.com/problem/binary-tree-level-order-traversal

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

Example Given binary tree {3,9,20,#,#,15,7},

return its level order traversal as:

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

Challenge: Using only 1 queue to implement it.

Solution: 参考九章答案，BFS模板级解法！

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
 
 
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Level order a list of lists of integer
     */
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        // write your code here
    	ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>> ();
    	if(root==null){
    		return res;
    	}
    	Queue<TreeNode> queue = new LinkedList<TreeNode>();
    	queue.offer(root);
    	TreeNode temp = new TreeNode(0); 
    	
    	while(!queue.isEmpty()){
    		ArrayList<Integer> level = new ArrayList<Integer>();
    		int size = queue.size();
    		for(int i = 0; i < size; i++){
    			temp = queue.poll();
    			level.add(temp.val);
    			
    			if(temp.left!=null){
    				queue.offer(temp.left);
    			}
    			if(temp.right!=null){
    				queue.offer(temp.right);
    			}   			  			
    		}
    		res.add(level);    		
    	}   	
    	return res;
    }
}