Leetcode 107-Binary Tree Level Order Traversal II(二叉树的层次遍历II)

Leetcode 107-Binary Tree Level Order Traversal II(二叉树的层次遍历II)

题目描述

给定一个二叉树,返回其节点值自底向上的层次遍历。 (即按从叶子节点所在层到根节点所在的层,逐层从左向右遍历)

例如:
给定二叉树 [3,9,20,null,null,15,7],

    3
   / \
  9  20
     /  \
   15   7

返回其自底向上的层次遍历为:

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

解法一:使用队列进行BFS 、使用 栈逆序

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    // 解法一:使用栈逆序
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        Stack<List<Integer>> stack = new Stack<List<Integer>>();
        if(root == null){
            return res;
        }
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.add(root);
        while(!q.isEmpty()){
            List<Integer> levelList = new ArrayList<Integer>();
            int size = q.size();
            while(size != 0){
                TreeNode node = q.remove();
                if(node.left != null){
                    q.add(node.left);    
                }
                if(node.right != null){
                    q.add(node.right);    
                }
                levelList.add(node.val);
                size --;
            }
            stack.add(levelList);
        }
        while(!stack.isEmpty()){
            res.add(stack.pop());
        }
        return res;
    }
}

解法二:Queue、BFS、Collections.reverse(res)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
        
    // 解法二:使用Collections.reverse(res)逆序
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if(root == null){
            return res;
        }
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.add(root);
        while(!q.isEmpty()){
            List<Integer> levelList = new ArrayList<Integer>();
            int size = q.size();
            while(size != 0){
                TreeNode node = q.remove();
                levelList.add(node.val);
                if(node.left != null){
                    q.add(node.left);    
                }
                if(node.right != null){
                    q.add(node.right);    
                }
                size --;
            }
            res.add(levelList);
        }
        Collections.reverse(res);
        return res;
    }
}

猜你喜欢

转载自blog.csdn.net/PerryJennings/article/details/83213164