Wednesday, January 7, 2015

[LeetCode]Binary Zigzag Level Order Traversal



偶数层从左往右,奇数层从右往左,最后注意一下children的顺序就行。

代码如下:

iterative
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if (root == null)
return res;
int level = 0;
LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
while (true) {
res.add(new ArrayList<Integer>());
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode temp;
if (level % 2 == 0) {
temp = queue.removeFirst();
if (temp.left != null)
queue.addLast(temp.left);
if (temp.right != null)
queue.addLast(temp.right);
} else {
temp = queue.removeLast();
if (temp.right != null)
queue.addFirst(temp.right);
if (temp.left != null)
queue.addFirst(temp.left);
}
res.get(res.size() - 1).add(temp.val);
}
level++;
if (queue.isEmpty())
break;
}
return res;
}
}



recursive
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res = new LinkedList<List<Integer>>();
dfs(res, root, 1);
return res;
}
public void dfs(List<List<Integer>> res, TreeNode node, int level) {
if(node == null)
return;
if (res.size() < level)
res.add(new LinkedList<Integer>());
if (level % 2 != 0)
res.get(level - 1).add(res.get(level - 1).size(), node.val);
else
res.get(level - 1).add(0, node.val);
dfs(res, node.left, level + 1);
dfs(res, node.right, level + 1);
}
}

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