Binary Tree ZigZag Level Order Traversal leetcode java
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,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
题解:
这题同样是BFS,用一个flag记录是否需要reverse,如果需要的话就把reverse的结果存储即可。
代码如下:
1 public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
2 ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
3
4 if(root==null)
5 return res;
6
7 LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
8 queue.add(root);
9
int num = 0;
boolean reverse = false;//a flag
while(!queue.isEmpty()){
num = queue.size();
ArrayList<Integer> levelres = new ArrayList<Integer>();
for(int i = 0; i<num; i++){
TreeNode node = queue.poll();
levelres.add(node.val);
if(node.left!=null)
queue.add(node.left);
if(node.right!=null)
queue.add(node.right);
}
if(reverse){
Collections.reverse(levelres);
reverse = false;
}else{
reverse = true;
}
res.add(levelres);
}
return res;
}
1 public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
2 ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
3 if(root == null)
4 return res;
5
6 LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
7 queue.add(root);
8 Boolean reverse = false;
9 int nextlevel = 0;
int currlevel = 1;
ArrayList<Integer> tmp = new ArrayList<Integer>();
while(!queue.isEmpty()){
TreeNode t = queue.poll();
tmp.add(t.val);
currlevel--;
if(t.left!=null){
queue.add(t.left);
nextlevel++;
}
if(t.right!=null){
queue.add(t.right);
nextlevel++;
}
if(currlevel == 0){
currlevel = nextlevel;
nextlevel = 0;
if(reverse){
Collections.reverse(tmp);
reverse = false;
}else{
reverse = true;
}
res.add(tmp);
tmp = new ArrayList<Integer>();
}
}
return res;
}
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