Binary Tree Zigzag Level Order Traversal

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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,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its zigzag level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

我的方法是用了一个queue，一个stack ，用stack 对需要reverse 的level 进行处理

别人的方法是用了两个queue

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {        List<List<Integer>> res = new ArrayList<List<Integer>>();        if (root == null) {            return res;        }        Queue<TreeNode> queue = new LinkedList<TreeNode>();        int level = 0;        int levelSize;        queue.offer(root);        Stack<TreeNode> st = new Stack<TreeNode>();        while (!queue.isEmpty()) {            ArrayList<Integer> list = new ArrayList<Integer>();            levelSize = queue.size();            for (int i = 0; i < levelSize; i++) {                TreeNode cur = queue.poll();                if(cur.left != null) {                    queue.offer(cur.left);                }                if(cur.right != null) {                    queue.offer(cur.right);                }                if (level % 2 == 0) {                    list.add(cur.val);                } else {//如果是要reverse 的level ，则用一个stack 处理一下                    st.push(cur);                }            }              while (!st.empty()) {                list.add(st.pop().val);            }            res.add(new ArrayList<Integer>(list));            level++;          }        return res;    }}

别人的

/** * Copyright: NineChapter * - Algorithm Course, Mock Interview, Interview Questions * - More details on: http://www.ninechapter.com/ */public class Solution {    public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();        if (root == null) {            return result;        }        Queue<TreeNode> queue = new LinkedList<TreeNode>();        queue.offer(root);        int currLevelNodeNum = 1;        int nextLevelNodeNum = 0;        while (currLevelNodeNum != 0) {            ArrayList<Integer> currLevelResult = new ArrayList<Integer>();            nextLevelNodeNum = 0;            while (currLevelNodeNum != 0) {                TreeNode node = queue.poll();                currLevelNodeNum--;                currLevelResult.add(node.val);                if (node.left != null) {                    queue.offer(node.left);                    nextLevelNodeNum++;                }                if (node.right != null) {                    queue.offer(node.right);                    nextLevelNodeNum++;                }            }            result.add(0, currLevelResult);            currLevelNodeNum = nextLevelNodeNum;        }        return result;    }}

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