Leetcode 103. Binary Tree Zigzag Level Order Traversal

方法1: 这题和107题完全一模一样的题。第一种方法 dfs-preoder-recursion。时间复杂n，空间复杂h。

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

方法2： dfs-preorder-iteration。时间复杂n，空间复杂h。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null) return res;
        Stack<TreeNode> nodes = new Stack<>();
        Stack<Integer> levels = new Stack<>();
        nodes.push(root);
        levels.push(0);
        while(!nodes.isEmpty()){
            TreeNode node = nodes.pop();
            int level = levels.pop();
            if(level == res.size()){
                res.add(new ArrayList<Integer>());
                res.get(level).add(node.val);
            }else{
                if(level % 2 == 0){
                    res.get(level).add(node.val);
                }else{
                    res.get(level).add(0, node.val);
                }
            }
            if(node.right != null){
                nodes.push(node.right);
                levels.push(level + 1);
            }
            if(node.left != null){
                nodes.push(node.left);
                levels.push(level + 1); 
            }  
        }
        return res;
    }
}

方法3: bfs-iteration。时间复杂n，空间复杂n。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null) return res;
        Queue<TreeNode> currLevel = new LinkedList<>();
        Queue<TreeNode> nextLevel = new LinkedList<>();
        nextLevel.add(root);
        while(!nextLevel.isEmpty()){
            currLevel = new LinkedList<>(nextLevel);
            nextLevel.clear();
            List<Integer> list = new LinkedList<Integer>();
            res.add(list);
            for(TreeNode node : currLevel){
                if(res.size() % 2 == 0){
                    list.add(0, node.val);
                }else{
                    list.add(node.val);
                }
                if(node.left != null)nextLevel.add(node.left);
                if(node.right != null)nextLevel.add(node.right);
            }
        }
        return res;
    }
}

总结：

• 无