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[LeetCode]Binary Tree Zigzag Level Order Traversal

2015-12-04 09:25 447 查看
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if (root == null) {
return result;
}
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
boolean flg = true;
while (!stack.isEmpty()) {
List<Integer> list = new ArrayList<Integer>();
Queue<TreeNode> queue = new LinkedList<TreeNode>();
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
list.add(node.val);
if (flg) {
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
} else {
if (node.right != null) {
queue.offer(node.right);
}
if (node.left != null) {
queue.offer(node.left);
}
}
}
while (!queue.isEmpty()) {
stack.push(queue.poll());
}
flg = !flg;
result.add(list);
}
return result;
}
}


下面的方法代码复杂度低一些

public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if (root == null) {
return result;
}
boolean flg = false;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
while (!queue.isEmpty()) {
List<Integer> list = new LinkedList<Integer>();
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
if (flg) {
list.add(0, node.val);
} else {
list.add(node.val);
}
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
result.add(list);
flg = !flg;
}
return result;
}
}
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