层级遍历_leetcode十题合集

层次遍历本质上是BFS，维护一个队列，把根节点加入队列，每一个节点出队的时候把它的子节点加入队列直到队列中没有节点。

1.leetcode102二叉树的层次遍历

public List<List<Integer>> levelOrder(TreeNode root) {
    List<List<Integer>> res = new ArrayList();
    if(root == null) return res;
    Deque<TreeNode> q = new LinkedList();
    q.offer(root);
    while(!q.isEmpty()) {
        int size = q.size();
        List<Integer> list = new ArrayList();
        while(size > 0) {
            TreeNode treeNode = q.poll();
            list.add(treeNode.val);
            if(treeNode.left != null) q.offer(treeNode.left);
            if(treeNode.right != null) q.offer(treeNode.right);
            size--;
        }
        res.add(list);
    }
    return res;
}

2.leetcode107二叉树层次遍历II

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

3.leetcode199 二叉树的右视图

这个题本质上就是二叉树的层次遍历，只要保留每一层的最后一个节点就好了。
这告诉我们一个道理，要学会活学活用。

public List<Integer> rightSideView(TreeNode root) {
    List<Integer> list = new ArrayList();
    if(root == null) return list;
    Deque<TreeNode> q = new LinkedList();
    q.offer(root);
    while(!q.isEmpty()) {
        int size = q.size();
        for(int i = 0; i < size; i++) {
            TreeNode node = q.poll();
            if(node.left != null) q.offer(node.left);
            if(node.right != null) q.offer(node.right);
            if(i == size-1) list.add(node.val);
        }
    }
    return list;
}

插一句题外话，我看到了三年前自己的提交记录，2019.3.12的下午三点到五点，我用了两个小时把这题做出来。现在我能直接bug-free AK。进步了，但这个进步速度太慢了。

4.leetcode637 二叉树的层平均值

public List<Double> averageOfLevels(TreeNode root) {
    List<Double> list = new ArrayList();
    if(root == null) return list;
    Deque<TreeNode> q = new LinkedList();
    q.offer(root);
    while(!q.isEmpty()) {
        int size = q.size();
        double sum = 0;
        for(int i = 0; i < size; i++) {
            TreeNode node = q.poll();
            if(node.left != null) q.offer(node.left);
            if(node.right != null) q.offer(node.right);
            sum += node.val;
        }
        list.add(sum /size);
    }
    return list;
}

5.leetcoder429 N叉树的层次遍历

public List<List<Integer>> levelOrder(Node root) {
    List<List<Integer>> res = new ArrayList();
    if(root == null) return res;
    Deque<Node> q = new LinkedList();
    q.offer(root);
    while(!q.isEmpty()) {
        int size = q.size();
        List<Integer> list = new ArrayList();
        for(int i = 0; i < size; i++) {
            Node node = q.poll();
            for(Node n: node.children) {
                if(n != null) q.offer(n);
            }
            list.add(node.val);
        }
        res.add(list);
    }
    return res;
}

6.leetcode515 树的每行中的最大值

public List<Integer> largestValues(TreeNode root) {
    List<Integer> list = new ArrayList();
    if(root == null) return list;
    Deque<TreeNode> q = new LinkedList();
    q.offer(root);
    while(!q.isEmpty()) {
        int size = q.size();
        int ans = Integer.MIN_VALUE;
        for(int i = 0; i < size; i++) {
            TreeNode node = q.poll();
            if(node.left != null) q.offer(node.left);
            if(node.right != null) q.offer(node.right);
            ans = Math.max(ans, node.val);
        }
        list.add(ans);
    }
    return list;
}

7.leetcode116 填充每个节点的下一个右侧节点指针I

public Node connect(Node root) {
    Deque<Node> q = new LinkedList();
    if(root == null) return root;
    q.offer(root);
    while(!q.isEmpty()) {
        int size = q.size();
        for(int i = 0; i < size; i++) {
            Node node1 = q.poll();
            Node node2 = q.peek();
            if(node1.left != null) q.offer(node1.left);
            if(node1.right != null) q.offer(node1.right);
            if(i != size-1) node1.next = node2;
            else node1.next = null;
        }
    }
    return root;
}

8.leetcode117 填充每个节点的下一个右侧节点指针II

public Node connect(Node root) {
    if(root == null) return root;
    Deque<Node> q = new LinkedList();
    q.offer(root);
    while(!q.isEmpty()) {
        int size = q.size();
        for(int i = 0; i < size; i++) {
            Node node1 = q.poll();
            if(node1.left != null) q.offer(node1.left);
            if(node1.right != null) q.offer(node1.right);
            if(i != size -1) {
                Node node2 = q.peek();
                node1.next =node2;
            }
            else {
                node1.next = null;
            }
        }
    }
    return root;
}

9.leetcode104 二叉树的最大深度

public int maxDepth(TreeNode root) {
    int ans = 0;
    if(root == null) return 0;
    Deque<TreeNode> q = new LinkedList();
    q.offer(root);
    while(!q.isEmpty()) {
        int size = q.size();
        ans++;
        for(int i = 0; i < size; i++) {
            TreeNode node = q.poll();
            if(node.left != null) q.offer(node.left);
            if(node.right != null) q.offer(node.right);
        }
    }
    return ans;
}

10.leetcode111 二叉树最小深度

public int minDepth(TreeNode root) {
    int depth = 0;
    if(root == null) return depth;
    Deque<TreeNode> q = new LinkedList();
    q.offer(root);
    while(!q.isEmpty()) {
        int size = q.size();
        depth++;
        for(int i = 0; i < size; i++) {
            TreeNode node = q.poll();
            if(node.left == null && node.right == null) return depth;
            if(node.left != null) q.offer(node.left);
            if(node.right != null) q.offer(node.right);
        }
    }
    return depth;
}