[LeetCode]102. Binary Tree Level Order Traversal

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https://leetcode.com/problems/binary-tree-level-order-traversal/

层序遍历二叉树

二叉树问题，很多递归调用时要传入当前位置的深度depth！！！！

解法一：

非递归，注意：1、queue的API；2、只用一个queue就可以完成

public class Solution {    public List<List<Integer>> levelOrder(TreeNode root) {        List<List<Integer>> res = new LinkedList();        if (root == null) {            return res;        }        Queue<TreeNode> queue = new LinkedList();        queue.offer(root);        while (!queue.isEmpty()) {            List<Integer> intList = new LinkedList();            int size = queue.size();            for (int i = 0; i < size; i++) {                if (queue.peek().left != null) {                    queue.add(queue.peek().left);                }                if (queue.peek().right != null) {                    queue.add(queue.peek().right);                }                intList.add(queue.poll().val);            }            res.add(intList);        }        return res;    }}

解法二：

递归，把当前遍历到的加入到当前的深度所代表的index的list中

public class Solution {    public List<List<Integer>> levelOrder(TreeNode root) {        List<List<Integer>> res = new LinkedList();        if (root == null) {            return res;        }        dfs(root, res, 0);        return res;    }    private void dfs(TreeNode root, List<List<Integer>> res, int depth) {        if (root == null) {            return;        }        if (depth >= res.size()) {            res.add(new LinkedList());        }        res.get(depth).add(root.val);        dfs(root.left, res, depth + 1);        dfs(root.right, res, depth + 1);    }}

倒序层序遍历二叉树只要把intList插入到res的第一个位置即可

https://leetcode.com/problems/binary-tree-level-order-traversal-ii/

public class Solution {    public List<List<Integer>> levelOrderBottom(TreeNode root) {        List<List<Integer>> res = new LinkedList();        if (root == null) {            return res;        }        Queue<TreeNode> queue = new LinkedList();        queue.offer(root);        while (!queue.isEmpty()) {            List<Integer> intList = new LinkedList();            int size = queue.size();            for (int i = 0; i < size; i++) {                TreeNode node = queue.peek();                if (node.left != null) {                    queue.offer(node.left);                }                if (node.right != null) {                    queue.offer(node.right);                }                intList.add(queue.poll().val);            }            res.add(0, intList);        }        return res;    }}

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