Binary Tree Level Order Traversal

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Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

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

    3   / \  9  20    /  \   15   7

return its level order traversal as:

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

对树进行层次遍历,当然啦，第一想法是递归，

public class Solution {    public List<List<Integer>> levelOrder(TreeNode root) {        int depth = 1;        List<List<Integer>> lst = new ArrayList<List<Integer>>();        levelOrder(lst,depth,root);        return lst;    }    public void levelOrder(List<List<Integer>> lst,int depth,TreeNode root){        if(root == null){            return;        }else{            if(lst.size() < depth){                List<Integer> lsIn = new ArrayList<Integer>();                lst.add(lsIn);            }            List<Integer> ls = lst.get(depth - 1);            ls.add(root.val);            if(root.left != null){                 levelOrder(lst,depth+1,root.left);            }            if(root.right != null){                levelOrder(lst,depth+1,root.right);            }                   }            }}

战胜了37.78%，效率的确不高。于是我们修改一下代码，利用链表进行操作。

public class Solution {    public List<List<Integer>> levelOrder(TreeNode root) {int cnt = 0;int f = 0;int t = 1;List<List<Integer>> lsts = new ArrayList<List<Integer>>();List<Integer> lst = new ArrayList<Integer>();List<TreeNode> nodeList = new ArrayList<TreeNode>();nodeList.add(root);if(root == null){return lsts;}while(t != 0){            if(f == t){            t = cnt;            f = 0;            cnt = 0;            lsts.add(lst);            lst = new ArrayList<Integer>();            }            else{            TreeNode node = nodeList.get(0);            if(node != null){            f ++;            lst.add(node.val);            }            if(node.left != null){            cnt ++;            nodeList.add(node.left);            }            if(node.right != null){            cnt ++;            nodeList.add(node.right);            }            nodeList.remove(node);            }}    return lsts;    }}

可是效率真的太低了，只战胜了11.63%，究其原因是不断的删除元素的操作引起的。我们除去删除的操作。每次生成新的链表存储每层的节点。

public List<List<Integer>> levelOrder(TreeNode root) {int cnt = 0;int f = 0;int t = 1;List<List<Integer>> lsts = new ArrayList<List<Integer>>();List<Integer> lst = new ArrayList<Integer>();List<TreeNode> nodeList = new ArrayList<TreeNode>();nodeList.add(root);if(root == null){return lsts;}while(t != 0){            if(f == t){            t = cnt;            f = 0;            cnt = 0;            lsts.add(lst);            lst = new ArrayList<Integer>();            }            else{            TreeNode node = nodeList.get(0);            if(node != null){            f ++;            lst.add(node.val);            }            if(node.left != null){            cnt ++;            nodeList.add(node.left);            }            if(node.right != null){            cnt ++;            nodeList.add(node.right);            }            nodeList.remove(0);            }}    return lsts;    }

效率还是不高哦，战胜了37.78% 。我们该怎么优化呢？于是我们可以利用一个链表进行操作，由下标进行确定每层的节点。

public List<List<Integer>> levelOrder(TreeNode root) {int cnt = 0;int f = 0;int t = 1;int index; List<List<Integer>> lsts = new ArrayList<List<Integer>>();List<Integer> lst = new ArrayList<Integer>();List<TreeNode> nodeList = new ArrayList<TreeNode>();nodeList.add(root);if(root == null){return lsts;}index = 0;while(t != 0){            if(f == t){            t = cnt;            f = 0;            cnt = 0;            lsts.add(lst);            lst = new ArrayList<Integer>();            }            else{            TreeNode node = nodeList.get(index ++);            if(node != null){            f ++;            lst.add(node.val);            }            if(node.left != null){            cnt ++;            nodeList.add(node.left);            }            if(node.right != null){            cnt ++;            nodeList.add(node.right);            }            }}    return lsts;    }

此次战胜了87.36% 。

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