採用廣度優先遍歷（Breath First Search），使用佇列實現，一般模板如下：

102. Binary Tree Level Order Traversal

//迭代，藉助佇列實現
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        Deque<TreeNode> queue = new LinkedList<>();
        if (root != null) queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> temp = new ArrayList<>();
            for (int i  = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                temp.add(cur.val);
                if (cur.left != null) queue.offer(cur.left);
                if (cur.right != null) queue.offer(cur.right);
            }
            res.add(temp);
        }
        return res;
    }
}
 

也可以使用遞迴：

//遞迴
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        order(root, 0, res);
        return res;
    }
    
    void order(TreeNode node, int deep, List<List<Integer>> res) {
        if (node == null) return;
        deep++;
        //層級界定
        if (res.size() < deep) {
            List<Integer> temp = new ArrayList<>();
            res.add(temp);
        }
        res.get(deep - 1).add(node.val);
        order(node.left, deep, res);
        order(node.right, deep, res);
    }
}
 

既然明白了基本原理，接下來就是打十個

107. 二叉樹的層序遍歷 II

class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        Deque<TreeNode> queue = new LinkedList<>();
        if (root != null) queue.offer(root);
        while (!queue.isEmpty()) {
            List<Integer> temp = new ArrayList<>();
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                temp.add(cur.val);
                if (cur.left != null) queue.offer(cur.left);
                if (cur.right != null) queue.offer(cur.right);
            }
            res.add(temp);
        }
        Collections.reverse(res);
        return res;
    }
}
 

199. 二叉樹的右檢視

class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        Deque<TreeNode> queue = new LinkedList<>();
        if (root != null) queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                if (cur.left != null) queue.offer(cur.left);
                if (cur.right != null) queue.offer(cur.right);
                //該層最右
                if (i == size - 1) res.add(cur.val);
            }
        }
        return res;
    }
}

637. 二叉樹的層平均值

class Solution {
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> res = new ArrayList<>();
        Deque<TreeNode> queue = new LinkedList<>();
        if (root != null) queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            double levelSum  = 0.0;
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                if (cur.left != null) queue.offer(cur.left);
                if (cur.right != null) queue.offer(cur.right);
                levelSum += cur.val;
            }
            res.add(levelSum / size);
        }
        return res;
    }
}

429. N 叉樹的層序遍歷

class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> res = new ArrayList<>();
        Deque<Node> queue = new LinkedList<>();
        if (root != null) queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> temp = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                Node cur = queue.poll();
                for (Node child : cur.children) {
                    if (child != null) queue.offer(child);
                }
                temp.add(cur.val);
            }
            res.add(temp);
        }
        return res;
    }
}

515. 在每個樹行中找最大值

class Solution {
    public List<Integer> largestValues(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        Deque<TreeNode> queue = new LinkedList<>();
        if (root != null) queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            int max = Integer.MIN_VALUE;
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                max = max > cur.val ? max : cur.val;
                if (cur.left != null) queue.offer(cur.left);
                if (cur.right != null) queue.offer(cur.right);
            }
            res.add(max);
        }
        return res;
    }
}

116. 填充每個節點的下一個右側節點指標

class Solution {
    public Node connect(Node root) {
        Deque<Node> queue = new LinkedList<>();
        if (root != null) queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            Node pre = queue.peek(); //前驅結點
            for (int i = 0; i < size; i++) {
                Node cur = queue.poll();
                if (cur.left != null) queue.offer(cur.left);
                if (cur.right != null) queue.offer(cur.right);
                //非該層一個節點，將當前節點賦值給前驅節點的後繼。
                if (i != 0) {
                    pre.next = cur;
                    pre = cur;
                }
                //next預設值null
                //if (i == size - 1) cur.next = null;
            }
        }
        return root;
    }
}

117. 填充每個節點的下一個右側節點指標 II

 程式碼與上同

留兩個明天繼續打。

參考：程式碼隨想錄：programmercarl.com