LeetCode-Binary Tree Level Order Traversal

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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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

题意：返回一颗二叉树的层次遍历；

解法：对于层次遍历我们可以考虑利用队列来实现；对于每一层，我们遍历队列中的所有元素，同时将其子节点入队，这样把当前层的结点遍历完之后，队列中剩余的就是下一层的结点(如果存在的话)；知道队列为空停止；

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) return res;
        
        LinkedList<TreeNode> nodes = new LinkedList<>();
        nodes.add(root);
        int cnt = 1;
        while (!nodes.isEmpty()) {
            int cnt_node = cnt;
            cnt = 0;
            List<Integer> levelNode = new ArrayList<>();
            while (cnt_node-- > 0) {
                TreeNode temp = nodes.remove();
                levelNode.add(temp.val);
                if (temp.left  != null) { nodes.add(temp.left);  cnt++; }
                if (temp.right != null) { nodes.add(temp.right); cnt++; }
            }
            res.add(levelNode);
        }
        
        return res;
    }
}