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] ]
解法一: bfs
//bfs
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if(root==null) return res;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
boolean reverse = false;
while(!queue.isEmpty()){
List<Integer> layer = new ArrayList<>();
int size = queue.size();
for(int i = 0;i<size;i++){
TreeNode prev = queue.poll();
if(reverse){
layer.add(0,prev.val);
}
else{
layer.add(prev.val);
}
if(prev.left!=null){
queue.offer(prev.left);
}
if(prev.right!=null){
queue.offer(prev.right);
}
}
res.add(layer);
reverse = !reverse;
}
return res;
}
解法二:dfs
//dfs(先序遍历)
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
dfs(root,res,0);
return res;
}
private void dfs(TreeNode root,List<List<Integer>> res,int depth){
if(root==null) return;
if(depth==res.size()){
res.add(new ArrayList<Integer>());
}
List<Integer> layer = res.get(depth);
if(depth%2==0){
layer.add(root.val);
}
else{
layer.add(0,root.val);
}
dfs(root.left,res,depth+1);
dfs(root.right,res,depth+1);
}
个人觉得bfs方法比较直观,容易理解