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leetcode(103) binary tree zigzag level order traversal 解法：

admin 11月 29, 2020 0

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,null,null,15,7],

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  3
 / 
9  20
  /  
 15   7

return its zigzag level order traversal as:

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

解法：

这道题需要仔细一点，虽然思路很简单，但是有点绕。要实现之字形走位，因为相邻行的方向不同，考虑使用两个栈进行操作，并且两个栈的进栈顺序也不同，一个先进右节点，一个先进左节点，具体代码如下：

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class  {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> result = new LinkedList<List<Integer>>();
        if (root == null) {
            return result;
        }
        Stack<TreeNode> s1 = new Stack<TreeNode>();
        Stack<TreeNode> s2 = new Stack<TreeNode>();
        s1.push(root);
        while (!s1.empty() || !s2.empty()) {
            if (s2.empty()) {
                List<Integer> tmp = new ArrayList<Integer>();
                while (!s1.empty()) {
                    TreeNode now = s1.pop();
                    if (now.left != null) {
                        s2.push(now.left);
                    }
                    if (now.right != null) {
                        s2.push(now.right);
                    }
                    tmp.add(now.val);
                }
                result.add(tmp);
            } else {
                List<Integer> tmp = new ArrayList<Integer>();
                while (!s2.empty()) {
                    TreeNode now = s2.pop();
                    if (now.right != null) {
                        s1.push(now.right);
                    }
                    if (now.left != null) {
                        s1.push(now.left);
                    }
                    tmp.add(now.val);
                }
                result.add(tmp);
            }
        }
        return result;
    }
}

另一种思路：层序遍历，然后对于某些层在输出时翻转list即可：

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        int level = 1;
        Queue<TreeNode> q = new LinkedList<>();
        Stack<TreeNode> s = new Stack<>();
        q.offer(root);
        while(!q.isEmpty()){
            int size = q.size();
            List<Integer> tmplist = new ArrayList<>();
            for(int i = 0; i < size; i++){
                TreeNode tmp = q.poll();
                tmplist.add(tmp.val);
                if(tmp.left!=null){
                    q.offer(tmp.left);     
                } 
                if(tmp.right!=null){
                    q.offer(tmp.right);      
                }
            }
            if(level%2==0){
                for(int i = 0;i<tmplist.size()/2;i++){
                    int tmp = tmplist.get(i);
                    tmplist.set(i,tmplist.get(tmplist.size() - i - 1));
                    tmplist.set(tmplist.size() - i - 1,tmp);
                }
                res.add(tmplist);
            }
            else{
                res.add(tmplist);
            }
            level++;
        }
        return res;
    }
}
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