BST----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).

Have you met this question in a real interview? Yes

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]

]

Challenge

Challenge 1: Using only 1 queue to implement it.

Challenge 2: Use DFS algorithm to do it.

递归算法

Java queue Api

/**
* 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>> result = new ArrayList<ArrayList<Integer>>();
if(root == null) return result ;
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
while(!q.isEmpty()){
int qLen = q.size();
ArrayList<Integer> aList = new ArrayList <Integer>();
for(int i = 0; i<qLen; i++){
TreeNode node = q.poll();//出队
aList.add(node.val);
if(node.left!= null) q.offer(node.left);
if(node.right!=null) q.offer(node.right);//入队

}
result.add(aList);
}
return result ;

}
}