2627二叉树的层次遍历和之字顺序打印

26.二叉树的层次遍历

用列队，BFS，时间复杂度和空间复杂度都是O(N)

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param root TreeNode类 
     * @return int整型ArrayList>
     */
    public ArrayList<ArrayList<Integer>> levelOrder (TreeNode root) {
        // write code here
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        Queue<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        while(!q.isEmpty()){
            //一层
            ArrayList<Integer> row = new ArrayList<Integer>();
            int count = q.size();
            //这层的个数刚好是队列的长度
            while(count-- >0){
                TreeNode node = q.poll();
                row.add(node.val);
                if(node.left!=null) q.offer(node.left);
                if(node.right!=null) q.offer(node.right);
            }
            res.add(row);
        }
        return res;
    }
}
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27.二叉树之字打印

一开始是在offer左节点和右节点的顺序上做判断，测试了一下发现是错误的，第二个例子就错误了，后面就直接在row的数组中判断奇数=偶层，偶层把row数组反转一下，其余和上面上一题的层次遍历完全一致、时间复杂度和空间复杂度都是O(N)

import java.util.*;
import java.util.ArrayList;

/*
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
public class Solution {
    public ArrayList<ArrayList<Integer> > Print(TreeNode pRoot) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        Queue<TreeNode> q = new ArrayDeque<>();
        if(pRoot==null) return new ArrayList<ArrayList<Integer>>(0);
        q.offer(pRoot);
        int index = 0;
        while(!q.isEmpty()){
            ArrayList<Integer> row = new ArrayList<>();
            ArrayList<Integer> rever = new ArrayList<>();
            int count = q.size();
            index++;
            while(count-- >0){
                    TreeNode node = q.poll();
                    row.add(node.val);                 
                    if(node.left!=null) q.offer(node.left);//奇数层从右到左                   
                    if(node.right!=null) q.offer(node.right);
            } 
            if(index%2==0){
                for(int i = row.size()-1; i>=0;i--){
                    rever.add(row.get(i));
                }
                res.add(rever);
            }
            else res.add(row);
        }
        return res;
    }

}
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