LeetCode-637. Average of Levels in Binary Tree(求树的每一层的平均值)(四种写法)

LeetCode-637. Average of Levels in Binary Tree(求树的每一层的平均值)

• BFS(层次)
• DFS(前序和中序递归)
  

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题目链接

题目

BFS(层次)

很容易想到的解法就是层次遍历，每次处理一层，先得到队列中所有元素的个数，然后全部处理完，然后处理下一层:

    /**
     * BFS做法
     */
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> res = new ArrayList<>();
        Queue<TreeNode>queue = new LinkedList<>();
        queue.add(root);
        TreeNode top = null;
        double sum = 0;
        while(!queue.isEmpty()){
            int n = queue.size();
            sum = 0;
            for(int i = 0; i < n; i++){
                top = queue.poll();
                sum += top.val;
                if(top.left != null)queue.add(top.left);
                if(top.right != null)queue.add(top.right);
            }
            res.add(sum/n);
        }
        return res;
    }

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DFS(前序和中序递归)

递归的做法就是 记录一个层数level，用两个List保存合以及个数，每次遍历到这一层的时候：

• 或者是这一层第一次来，那就加上第一次val，次数变为1；
• 或者不是第一次，不是第一层就累加这一层的和以及个数；
• 因为前序是先根，再左，再右，所以可以通过层数level和list的大写来判断上面的情况；

前序:

 /**
     * DFS递归  + 前序
     * @param root
     * @return
     */
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> res = new ArrayList<>();
        List<Double> sum = new ArrayList<>();
        List<Integer> count = new ArrayList<>();

        pre(root, 0, sum, count);

        for (int i = 0; i < sum.size(); i++) {
            res.add(sum.get(i) / count.get(i));
        }
        return res;
    }

    private static void pre(TreeNode root, int level, List<Double> sum, List<Integer> count) {
        if (root == null) return;
        if (level < sum.size()) { //回去的
            sum.set(level, sum.get(level) + root.val);
            count.set(level, count.get(level) + 1);
        } else {  //新的高度
            sum.add(1.0*root.val); //添加第一个
            count.add(1); 
        }
        pre(root.left, level + 1, sum, count);
        pre(root.right, level + 1, sum, count);
    }

中序:
中序遍历和前序遍历不同的是，先左子树走到底，然后回来根，然后再右子树，所以我们要预先给list中添加一个默认值，添加一个0即可。后序也是:

      public List<Double> averageOfLevels(TreeNode root) {
        List<Double> res = new ArrayList<>();
        List<Double> sum = new ArrayList<>();
        List<Integer> count = new ArrayList<>();
        in(root, 0, sum, count);
        for (int i = 0; i < sum.size(); i++) {
            res.add(sum.get(i) / count.get(i));
        }
        return res;
    }

    private static void in(TreeNode root, int level, List<Double> sum, List<Integer> count) {
        if (root == null) return;
        if(level >= sum.size()){
            sum.add(0.0);
            count.add(0);
        }
        in(root.left, level + 1, sum, count);
        sum.set(level, sum.get(level) + root.val);
        count.set(level, count.get(level) + 1);
        in(root.right, level + 1, sum, count);
    }

后序：

 public List<Double> averageOfLevels(TreeNode root) {
        List<Double> res = new ArrayList<>();
        List<Double> sum = new ArrayList<>();
        List<Integer> count = new ArrayList<>();
        pos(root, 0, sum, count);
        for (int i = 0; i < sum.size(); i++) {
            res.add(sum.get(i) / count.get(i));
        }
        return res;
    }

    private static void pos(TreeNode root, int level, List<Double> sum, List<Integer> count) {
        if (root == null) return;
        if(level >= sum.size()){
            sum.add(0.0);
            count.add(0);
        }
        pos(root.left, level + 1, sum, count);
        pos(root.right, level + 1, sum, count);
        sum.set(level, sum.get(level) + root.val);
        count.set(level, count.get(level) + 1);
    }