102. Binary Tree Level Order Traversal

2019/11/15 Leetcode

102. Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

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

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

Solution

方法1: BFS with Queue

图/树的层级遍历(level order traversal)是典型的BFS

public List<List<Integer>> levelOrder(TreeNode root) {
    List<List<Integer>> result = new ArrayList<>();
    if (root == null) return result;
    
    Queue<TreeNode> queue = new LinkedList<>();
    queue.offer(root);
    while (!queue.isEmpty()) {
        List<Integer> level = new ArrayList<>();
        int size = queue.size();
        for (int i = 0; i < size; i++) { // DON'T use size directly, it's changing
            TreeNode node = queue.poll();
            level.add(node.val);
            if (node.left != null) {
                queue.offer(node.left);
            }
            if (node.right != null) {
                queue.offer(node.right);
            }
        }
        result.add(level);
    }
    return result;
}

方法2:DFS Recursive

public List<List<Integer>> levelOrder(TreeNode root) {
    List<List<Integer>> res = new ArrayList<List<Integer>>();
    levelHelper(res, root, 0);
    return res;
}

public void levelHelper(List<List<Integer>> res, TreeNode root, int height) {
    if (root == null) return;
    if (height >= res.size()) {
        res.add(new LinkedList<Integer>());
    }
    res.get(height).add(root.val);
    levelHelper(res, root.left, height+1);
    levelHelper(res, root.right, height+1);
}

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