[LintCode] 69 Binary Tree Level Order Traversal 解题报告

Description
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},

    3
   / \
  9  20
      /  \
   15   7


return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]


思路
用队列来实现。size用来记录每一层有多少个元素。每一把拿完以后，队列里剩下的就是下一层所有的元素。


Code

/**
 * 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<>();
        if (root == null) {
            return result;
        }
        
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            ArrayList<Integer> level = new ArrayList<>();
            int size = queue.size();
            
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                level.add(node.val);
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            result.add(level);
        }
        return result;
    }
}