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BinaryTreeZigzagLevelOrderTraversal.java
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92 lines (80 loc) · 2.51 KB
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package oj.leetcode;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Stack;
/**
* 103. Binary Tree Zigzag Level Order Traversal
* Created by vonzhou on 2019/2/17.
*/
public class BinaryTreeZigzagLevelOrderTraversal {
/**
* AC
* DFS, 深度优先,遍历到对应层的时候把对应元素维护到结果集中
*
* @param root
* @return
*/
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
travel(root, res, 0);
return res;
}
private void travel(TreeNode curr, List<List<Integer>> sol, int level) {
if (curr == null) return;
if (sol.size() <= level) {
// 开辟保存新一层的结果
List<Integer> newLevel = new LinkedList<>();
sol.add(newLevel);
}
List<Integer> listOfThisLevel = sol.get(level);
// 偶数层的话从左到右, 奇数层从右到左
if (level % 2 == 0)
listOfThisLevel.add(curr.val);
else listOfThisLevel.add(0, curr.val);
travel(curr.left, sol, level + 1);
travel(curr.right, sol, level + 1);
}
/**
* AC
* 使用2个栈,每层的元素入同一个栈
* BFS,广度优先
*
* @param root
* @return
*/
public List<List<Integer>> zigzagLevelOrder2(TreeNode root) {
TreeNode p = root;
List<List<Integer>> res = new ArrayList<List<Integer>>();
if (p == null)
return res;
Stack<TreeNode> s1 = new Stack<TreeNode>();
Stack<TreeNode> s2 = new Stack<TreeNode>();
s1.push(root);
while (!s1.isEmpty() || !s2.isEmpty()) {
List<Integer> tmp = new ArrayList<Integer>();
while (!s1.isEmpty()) {
p = s1.pop();
tmp.add(p.val);
if (p.left != null)
s2.push(p.left);
if (p.right != null)
s2.push(p.right);
}
res.add(tmp);
tmp = new ArrayList<Integer>();
while (!s2.isEmpty()) {
p = s2.pop();
tmp.add(p.val);
// 奇数层,从右向左,所以先遍历右孩子
if (p.right != null)
s1.push(p.right);
if (p.left != null)
s1.push(p.left);
}
if (!tmp.isEmpty())
res.add(tmp);
}
return res;
}
}