102 Binary Tree Level Order Traversal

最新推荐文章于 2024-08-23 15:23:16 发布

dongbeier

最新推荐文章于 2024-08-23 15:23:16 发布

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本文介绍了一种经典的二叉树遍历算法——层次遍历，并提供了递归和迭代两种实现方式。通过这两种方法，可以有效地按层从左到右获取二叉树节点的值。

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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]
]

递归版

/**
 * 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 ArrayList<>();
        traversal(root, 0, res);
        return res;
    }
    
    private void traversal(TreeNode root, int level, List<List<Integer>> res) {
        if (root == null)
            return;
        if (level >= res.size())
            res.add(new ArrayList<>());
        res.get(level).add(root.val);
        traversal(root.left, level + 1, res);
        traversal(root.right, level + 1, res);
    }
}

迭代版

/**
 * 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 ArrayList<>();
        Queue<TreeNode> curr = new LinkedList<>();
        Queue<TreeNode> next = new LinkedList<>();
        
        if (root == null)
            return res;
        else
            curr.add(root);
        
        while (!curr.isEmpty()) {
            List<Integer> level = new ArrayList<>();
            while (!curr.isEmpty()) {
                TreeNode node = curr.poll();
                level.add(node.val);
                if (node.left != null) {
                    next.add(node.left);
                }
                
                if (node.right != null) {
                    next.add(node.right);
                }
            }
            res.add(level);
            Queue temp = curr;
            curr = next;
            next = temp;
        }
        return res;
    }
}