102. Binary Tree Level Order Traversal

Leetcode Tree Breath-first Search

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],

return its level order traversal as:

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

分析¶

这道题目要求的层序遍历。题目这么长，其实想让我们写广度优先搜索。

public List<List<Integer>> levelOrder(TreeNode root) {
    List<List<Integer>> levels = new ArrayList<List<Integer>>();
    if (root == null) return levels;
    Queue<TreeNode> queue = new LinkedList<>();
    queue.offer(root);
    while (!queue.isEmpty()) {
        List<Integer> curLevel = new ArrayList<>();
        int size = queue.size();
        for (int i = 0; i < size; i++) {
            TreeNode node = queue.poll();
            if (node.left != null) queue.offer(node.left);
            if (node.right != null) queue.offer(node.right);
            curLevel.add(node.val);
        }
        levels.add(curLevel);
    }
    return levels;    
}

不过也可以用深度优先搜索写

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);
}