subject

Given a binary tree, it returns the value of the nodes it traverses hierarchically. (that is, access all nodes layer by layer, from left to right).

For example:
Given a binary tree: [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

Return its hierarchical traversal result:

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

Title Solution

The sequence traversal taught in our data structure book is to use a queue to continuously put left and right subtrees into the queue. But this topic also requires output according to the layer. So the key question is: how to make sure it is on the same level.
We naturally think:
If all nodes of the previous layer are dequeued before joining the team, then these nodes of the dequeued are the list of the previous layer.
Since the queue is a first in, first out data structure, the list is left to right.

/**
 * 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 LinkedList<>();
        if (root == null) {
            return res;
        }

        LinkedList<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> currentRes = new LinkedList<>();
            // The size of the current queue is the number of nodes in the previous layer, and the queue will be out in turn
            while (size > 0) {
                TreeNode current = queue.poll();
                if (current == null) {
                    continue;
                }
                currentRes.add(current.val);
                // Left subtree and right subtree join the team
                if (current.left != null) {
                    queue.add(current.left);
                }
                if (current.right != null) {
                    queue.add(current.right);
                }
                size--;
            }
            res.add(currentRes);
        }
        return res;
    }
}

Can this problem be solved by non delivery?

Recursive subproblem: traversing the current node, for the current layer, traversing the next layer of the left subtree, traversing the next layer of the right subtree

Recursion end condition: current layer, current subtree node is null

/**
 * 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 LinkedList<>();
        if (root == null) {
            return res;
        }
        levelOrderHelper(res, root, 0);
        return res;
    }

    /**
     * @param depth The depth of binary tree
     */
    private void levelOrderHelper(List<List<Integer>> res, TreeNode root, int depth) {
        if (root == null) {
            return;
        }
        
        if (res.size() <= depth) {
            // The first node of the current layer needs a new list to store the current layer
            res.add(new LinkedList<>());
        }
        // depth layer, add the current node
        res.get(depth).add(root.val);
        // Recursively traverses the next level
        levelOrderHelper(res, root.left, depth + 1);
        levelOrderHelper(res, root.right, depth + 1);
    }
}

Popular reading

• Summary of technical articles
• [Leetcode] 101. Symmetric binary tree
• [Leetcode] 100. Same tree
• [Leetcode] 98. Verify binary search tree