Skimming records -- sequence traversal of binary tree

Catalog

Title Description ： Given a binary tree , Returns the result of its sequence traversal

Combined with auxiliary queue , Two layers of circulation , From top to bottom , Traverse from left to right , The result is sequence traversal

recursive ：

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Title Description ： Given a binary tree , Returns the result of its sequence traversal

Combined with auxiliary queue , Two layers of circulation , From top to bottom , Traverse from left to right , The result is sequence traversal

void levelTraverse(TreeNode root) {
    if (root == null) return;
    Queue<TreeNode> q = new LinkedList<>();
    q.offer(root);

    //  Traverse every layer of the binary tree from top to bottom 
    while (!q.isEmpty()) {
        int sz = q.size();
        //  Traverse each node of each layer from left to right 
        for (int i = 0; i < sz; i++) {
            TreeNode cur = q.poll();
            //  Put the next level node in the queue 
            if (cur.left != null) {
                q.offer(cur.left);
            }
            if (cur.right != null) {
                q.offer(cur.right);
            }
        }
    }
}

recursive ：

class Solution {
    List<List<Integer>> list = new ArrayList<>();
    public List<List<Integer>> levelOrder(TreeNode root) {
        dns(root,0);
        return list;
    }
    public void dns(TreeNode node,int lever){
        if(node == null) return;
        if(list.size()==lever) list.add(new ArrayList<Integer>());

        list.get(lever).add(node.val);

        dns(node.left,lever+1);
        dns(node.right,lever+1);
    }
}