LeetCode:199二叉树的右视图

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<>();

        if(root == null){
            return res;
        }

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while(!queue.isEmpty()){
            int size = queue.size();

            for(int i = 0; i < size; i++){
                TreeNode current = queue.poll();
                //这个判断条件代表刚好落在了该层最右边的元素上
                if(i == size - 1){
                    res.add(current.val);
                }
                //左右节点入队
                if(current.left != null){
                    queue.offer(current.left);
                }
                if(current.right != null){
                    queue.offer(current.right);
                }
            }
        }
        return res;
    }
}

思路：

计算每一层的节点数，然后循环将该层节点的最后一个存放进res即可。

如果求左视图：直接将判断条件改成if(i == 0)即可

如果求二叉树的层平均值：在for循环中加入sum += current.val,循环结束后用sum/size算平均分