94 lines
2.7 KiB
Java
94 lines
2.7 KiB
Java
//给定一个二叉树，返回其节点值的锯齿形层序遍历。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。
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//
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// 例如：
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//给定二叉树 [3,9,20,null,null,15,7],
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//
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//
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// 3
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// / \
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// 9 20
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// / \
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// 15 7
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//
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//
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// 返回锯齿形层序遍历如下：
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//
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//
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//[
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// [3],
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// [20,9],
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// [15,7]
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//]
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//
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// Related Topics 栈 树 广度优先搜索
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// 👍 421 👎 0
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package leetcode.editor.cn;
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import com.code.leet.entiy.TreeNode;
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import java.util.*;
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//103:二叉树的锯齿形层序遍历
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public class BinaryTreeZigzagLevelOrderTraversal {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new BinaryTreeZigzagLevelOrderTraversal().new Solution();
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List<Integer> list = Arrays.asList(3,9,20,null,null,15,7);
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solution.zigzagLevelOrder(new TreeNode(list));
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
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class Solution {
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public List<List<Integer>> zigzagLevelOrder(TreeNode root) {List<List<Integer>> ans = new LinkedList<List<Integer>>();
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if (root == null) {
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return ans;
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}
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Queue<TreeNode> nodeQueue = new LinkedList<TreeNode>();
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nodeQueue.offer(root);
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boolean isOrderLeft = true;
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while (!nodeQueue.isEmpty()) {
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Deque<Integer> levelList = new LinkedList<Integer>();
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int size = nodeQueue.size();
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for (int i = 0; i < size; ++i) {
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TreeNode curNode = nodeQueue.poll();
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if (isOrderLeft) {
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levelList.offerLast(curNode.val);
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} else {
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levelList.offerFirst(curNode.val);
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}
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if (curNode.left != null) {
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nodeQueue.offer(curNode.left);
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}
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if (curNode.right != null) {
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nodeQueue.offer(curNode.right);
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}
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}
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ans.add(new LinkedList<Integer>(levelList));
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isOrderLeft = !isOrderLeft;
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}
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return ans;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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} |