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[LeetCode – BFS & Stack] 103. Binary Tree Zigzag Level Order Traversal

编程语言 jitianyu123 34℃ 0评论
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1.1 题目

Given a binary tree, return the zigzag level order traversal of its nodes’ values. (ie, from left to right, then right to left for the next level and alternate between).


For example:


Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its zigzag level order traversal as:

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

2.2 分析

我的思路是修改树的BFS算法,使用两个栈,一个从左到右,一个从右到左存储遍历到的节点, 当节点入栈的同时,将节点中的数值存储到对应的结果数组中。


时间复杂度是O(n),但是空间复杂度是O(2n)

3.3 代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List> zigzagLevelOrder(TreeNode root) {

        List> result = new ArrayList<>();
        List tmpRes = new ArrayList<>();

        LinkedList L2R = new LinkedList<>();
        LinkedList R2L = new LinkedList<>();

        if(root == null){
            return result;
        }

        L2R.push(root);
        result.add(new ArrayList(Arrays.asList(root.val)));

        while(L2R.size() != 0 || R2L.size() != 0){
            tmpRes.clear();
            while(L2R.size() != 0){
                TreeNode cur = L2R.pop();
                if(cur.right != null){
                    R2L.push(cur.right);
                    tmpRes.add(cur.right.val);                    
                }
                if(cur.left != null){
                    R2L.push(cur.left);
                    tmpRes.add(cur.left.val);                    
                }
            }
            if(tmpRes.size() != 0){
                result.add(new ArrayList(tmpRes));
                tmpRes.clear();
            }

            while(R2L.size() != 0){
                TreeNode cur = R2L.pop();
                if(cur.left != null){
                    L2R.push(cur.left);
                    tmpRes.add(cur.left.val);                    
                }
                if(cur.right != null){
                    L2R.push(cur.right);
                    tmpRes.add(cur.right.val);                    
                }
            }
            if(tmpRes.size() != 0){
                result.add(new ArrayList(tmpRes));
                tmpRes.clear();
            }
        }

        return result;
    }
}

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