leetcode--Binary Tree Zigzag Level Order Traversal
2017-08-08 11:39
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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
return its zigzag level order traversal as:
题意:给定二叉树,遍历每一层。对于当前层,如果从左到右,则下一层从右到左。
分类:二叉树
解法1:层次遍历。使用一个标记来标记层的结束。每次结束将队列里面的数据保存。
[java] view
plain copy
/**
* 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<List<Integer>> zigzagLevelOrder(TreeNode root) {
Stack<TreeNode> stack = new Stack<TreeNode>();
LinkedList<List<Integer>> res = new LinkedList<List<Integer>>();
if(root == null) return res;
List<Integer> t = new ArrayList<Integer>();
int low = 0;
int high = 0;
int ceng = 0;
boolean flag=true;
stack.add(root);
t.add(root.val);
res.add(t);
t = new ArrayList<Integer>();;
while(low<=high){
TreeNode cur = stack.get(low);
if(cur.left!=null){
stack.add(cur.left);
t.add(cur.left.val);
high++;
}
if(cur.right!=null){
stack.add(cur.right);
t.add(cur.right.val);
high++;
}
if(low==ceng){
if(flag){
Collections.reverse(t);
}
flag = !flag;
if(t.size()!=0) res.add(t);;
t = new ArrayList<Integer>();
ceng = high;
}
low++;
}
return res;
}
}
解法2:层次遍历。使用LinkedList从而可以在头部添加节点。
使用level来做标记从而省去层次标记。
[java] view
plain copy
/**
* 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<List<Integer>> zigzagLevelOrder(TreeNode root) {
LinkedList<TreeNode> queue = new LinkedList<TreeNode>();//队列,用于层次遍历
List<List<Integer>> res = new ArrayList<List<Integer>>();//结果
List<Integer> cur = new ArrayList<Integer>();//保留当前层数据
if(root==null) return res;
boolean lToR = true;//是否从左到右
int level = 1;
queue.add(root);
while(queue.size()>0){
TreeNode node = queue.poll();
if(lToR)
cur.add(node.val);
else
cur.add(0,node.val);
if(node.left != null)
queue.add(node.left);
if(node.right != null)
queue.add(node.right);
if(--level == 0){
level = queue.size();
res.add(new ArrayList(cur));
cur.clear();
lToR = !lToR;
}
}
return res;
}
}
原文链接http://blog.csdn.net/crazy__chen/article/details/46484543
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] ]
题意:给定二叉树,遍历每一层。对于当前层,如果从左到右,则下一层从右到左。
分类:二叉树
解法1:层次遍历。使用一个标记来标记层的结束。每次结束将队列里面的数据保存。
[java] view
plain copy
/**
* 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<List<Integer>> zigzagLevelOrder(TreeNode root) {
Stack<TreeNode> stack = new Stack<TreeNode>();
LinkedList<List<Integer>> res = new LinkedList<List<Integer>>();
if(root == null) return res;
List<Integer> t = new ArrayList<Integer>();
int low = 0;
int high = 0;
int ceng = 0;
boolean flag=true;
stack.add(root);
t.add(root.val);
res.add(t);
t = new ArrayList<Integer>();;
while(low<=high){
TreeNode cur = stack.get(low);
if(cur.left!=null){
stack.add(cur.left);
t.add(cur.left.val);
high++;
}
if(cur.right!=null){
stack.add(cur.right);
t.add(cur.right.val);
high++;
}
if(low==ceng){
if(flag){
Collections.reverse(t);
}
flag = !flag;
if(t.size()!=0) res.add(t);;
t = new ArrayList<Integer>();
ceng = high;
}
low++;
}
return res;
}
}
解法2:层次遍历。使用LinkedList从而可以在头部添加节点。
使用level来做标记从而省去层次标记。
[java] view
plain copy
/**
* 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<List<Integer>> zigzagLevelOrder(TreeNode root) {
LinkedList<TreeNode> queue = new LinkedList<TreeNode>();//队列,用于层次遍历
List<List<Integer>> res = new ArrayList<List<Integer>>();//结果
List<Integer> cur = new ArrayList<Integer>();//保留当前层数据
if(root==null) return res;
boolean lToR = true;//是否从左到右
int level = 1;
queue.add(root);
while(queue.size()>0){
TreeNode node = queue.poll();
if(lToR)
cur.add(node.val);
else
cur.add(0,node.val);
if(node.left != null)
queue.add(node.left);
if(node.right != null)
queue.add(node.right);
if(--level == 0){
level = queue.size();
res.add(new ArrayList(cur));
cur.clear();
lToR = !lToR;
}
}
return res;
}
}
原文链接http://blog.csdn.net/crazy__chen/article/details/46484543
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