标签：层次遍历 new int init empty return -- 研究 遍历
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] ]
层次遍历+ 逆序偶数行
/**
* 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) {
List<List<Integer>> ans = new ArrayList<>();
if (root == null) {
//ans.add(list);
return ans;
}
Queue<TreeNode> q = new LinkedList<>();
int deep = 0;
q.offer(root);
while (!q.isEmpty()) {
int size = q.size();
deep++;
List<Integer> list = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode cur = q.poll();
list.add(cur.val);
if (cur.left != null) {
q.offer(cur.left);
}
if (cur.right != null) {
q.offer(cur.right);
}
}
if (deep % 2 == 1) {
ans.add(list);
} else {
reverse(list);
ans.add(list);
}
}
return ans;
}
private void reverse(List<Integer> list) {
int i = 0, j = list.size() - 1;
while (i < j) {
int temp = (int)list.get(i);
list.set(i, list.get(j));
list.set(j, temp);
i++;
j--;
}
}
}
贴一个DFS做法：很好, 但是还是bfs 好想, 以后再研究
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root)
{
List<List<Integer>> sol = new ArrayList<>();
travel(root, sol, 0);
return sol;
}
private void travel(TreeNode curr, List<List<Integer>> sol, int level)
{
if(curr == null) return;
if(sol.size() <= level)
{
List<Integer> newLevel = new LinkedList<>();
sol.add(newLevel);
}
List<Integer> collection = sol.get(level);
if(level % 2 == 0) collection.add(curr.val);
else collection.add(0, curr.val);
travel(curr.left, sol, level + 1);
travel(curr.right, sol, level + 1);
}
}
103. Binary Tree Zigzag Level Order Traversal
标签：层次遍历 new int init empty return -- 研究 遍历
原文地址：http://www.cnblogs.com/apanda009/p/7296558.html
