Leetcode 102. Binary Tree Level Order Traversal

最新推荐文章于 2023-11-06 11:13:49 发布

原创最新推荐文章于 2023-11-06 11:13:49 发布 · 121 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#tree #dfs #bfs #recurison #iteration

Leetcode 专栏收录该内容

278 篇文章

订阅专栏

本文对比了三种实现二叉树层次遍历的方法：递归深度优先（DFS）、迭代双栈（DFS-Preorder-Iteration）和广度优先（BFS-Iteration-Queues）。它们都以时间复杂度O(n)和空间复杂度O(n)为特点，适用于查找二叉树节点的层次顺序。

在这里插入图片描述
方法1: 这道题和107题一模一样，可以一起看。方法1是dfs-preorder-recursion。时间复杂n，空间复杂n。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    List<List<Integer>> res = new ArrayList<>();
    public List<List<Integer>> levelOrder(TreeNode root) {
        if(root == null) return res;
        dfsPreorder(root, 0);
        return res;
    }
    
    public void dfsPreorder(TreeNode root, int level){
        if(level == res.size()) res.add(new ArrayList<Integer>());
        res.get(level).add(root.val);
        if(root.left != null) dfsPreorder(root.left, level + 1);
        if(root.right != null) dfsPreorder(root.right, level + 1);
    }
}

方法2: dfs-preorder-iteration-two stacks。时间复杂n，空间复杂h。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null) return res;
        Stack<TreeNode> nodes = new Stack<>();
        Stack<Integer> levels = new Stack<>();
        nodes.push(root);
        levels.push(0);
        while(!nodes.isEmpty()){
            TreeNode node = nodes.pop();
            int level = levels.pop();
            if(level == res.size()) res.add(new ArrayList<Integer>());
            res.get(level).add(node.val);
            if(node.right != null){
                nodes.push(node.right);
                levels.push(level + 1);
            }
            if(node.left != null){
                nodes.push(node.left);
                levels.push(level + 1);
            }  
        }
        return res;
    }
}

方法3: bfs-iteration-two queues。时间复杂n，空间复杂n。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null) return res;
        Queue<TreeNode> currLevel = new LinkedList<>();
        Queue<TreeNode> nextLevel = new LinkedList<>();
        nextLevel.offer(root);
        while(!nextLevel.isEmpty()){
            currLevel = new LinkedList<>(nextLevel);
            nextLevel.clear();
            res.add(new ArrayList<>());
            for(TreeNode node : currLevel){
                res.get(res.size()-1).add(node.val);
                if(node.left != null) nextLevel.offer(node.left);
                if(node.right != null) nextLevel.offer(node.right);
            }
        }
        return res;
    }
}

总结：