103. Binary Tree Zigzag Level Order Traversal

题目：

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,#,#,15,7},

    3
   / 
  9  20
    /  
   15   7

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]

链接： http://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/

题解：

跟level order traversal一样，只不过多了一个flag来判断添加元素的顺序。

Time Complexity - O(n)， Space Complexity - O(n).

/**
 * Definition for a binary tree node.
 * 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<>();
        if(root == null)
            return res;
        ArrayList<Integer> list = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        int curLevel = 1, nextLevel = 0;
        boolean zigZagFlag = true;
        
        while(!q.isEmpty()) {
            TreeNode node = q.poll();
            if(zigZagFlag)
                list.add(node.val);
            else
                list.add(0, node.val);
            curLevel--;
            if(node.left != null) {
                q.offer(node.left);
                nextLevel++;
            }
            if(node.right != null) {
                q.offer(node.right);
                nextLevel++;
            }
            if(curLevel == 0) {
                curLevel = nextLevel;
                nextLevel = 0;
                res.add(new ArrayList<Integer>(list));
                list.clear();
                zigZagFlag = !zigZagFlag;
            }
        }
        
        return res;
    }
}

二刷:

注意要写得流畅。   有一个boolean变量 zigzag来确定合适按照正序 / 逆序 添加结果到每一个level的list里。

Java:

/**
 * Definition for a binary tree node.
 * 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<>();
        if (root == null) return res;
        List<Integer> list = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        int curLevel = 1, nextLevel = 0;
        boolean zigzag = true;
        
        while (!q.isEmpty()) {
            TreeNode node = q.poll();
            if (zigzag) list.add(node.val);
            else list.add(0, node.val);
            curLevel--;
            if (node.left != null) {
                q.offer(node.left);
                nextLevel++;
            }
            if (node.right != null) {
                q.offer(node.right);
                nextLevel++;
            }
            if (curLevel == 0) {
                curLevel = nextLevel;
                nextLevel = 0;
                res.add(new ArrayList<>(list));
                list.clear();
                zigzag = !zigzag;
            }
        }
        return res;
    }
}

测试：