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102. 二叉树的层序遍历

创建一个队列，保存存进去的结点，每次记录队列大小，来进行循环出结点，放下一层结点。

/**
 * 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>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if(root == null) {
            return result;
        }
        Deque<TreeNode> deque = new LinkedList<>();
        deque.offer(root);

        while(!deque.isEmpty()) {
            int size = deque.size();
            List<Integer> list = new ArrayList<>();

            for(int i = 0; i < size; i++) {
                TreeNode cur = deque.poll();
                list.add(cur.val);
                if(cur.left != null) {
                    deque.offer(cur.left);
                }
                if(cur.right != null) {
                    deque.offer(cur.right);
                }
            }
            result.add(list);
        }
        return result;
    }
}

107.二叉树的层次遍历 II

题目链接：二叉树的层次遍历 II

要求是从底层往上存。
第一种思路就是按照原来的顺序，最后再把结果翻转即可
第二种思路就是仍然循环插入，但是每次直接把list插在头部就行了
在这里我想使用一下第二种方法，所以原先的List就用不了了，可以用Linkelist

/**
 * 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>> levelOrderBottom(TreeNode root) {
        LinkedList<List<Integer>> result = new LinkedList<>();
        if(root == null) {
            return result;
        }
        Deque<TreeNode> deque = new LinkedList<>();
        deque.offer(root);

        while(!deque.isEmpty()) {
            int size = deque.size();
            List<Integer> list = new ArrayList<>();

            for(int i = 0; i < size; i++) {
                TreeNode cur = deque.poll();
                list.add(cur.val);
                if(cur.left != null) {
                    deque.add(cur.left);
                }
                if(cur.right != null) {
                    deque.add(cur.right);
                }
            }
            result.addFirst(list);
        }
        return result;
    }
}

199. 二叉树的右视图

题目链接：二叉树的右视图

当遍历到某一层的最后一个结点，就把它的值取出来放进result。
如何判断到了最后一个结点呢。就是当每一层循环到i == size -1 时。

/**
 * 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<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        Deque<TreeNode> deque = new LinkedList<>();

        if(root == null) {
            return result;
        }
        deque.offer(root);

        while(!deque.isEmpty()) {
            int size = deque.size();
            for(int i = 0; i < size; i++) {
                TreeNode cur = deque.poll();
                if(cur.left != null) {
                    deque.offer(cur.left);
                }
                if(cur.right != null) {
                    deque.offer(cur.right);
                }
                if(i == size - 1) {
                    result.add(cur.val);
                }
            }
        }
        return result;
    }
}

637. 二叉树的层平均值

题目链接：二叉树的层平均值

/**
 * 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<Double> averageOfLevels(TreeNode root) {
        List<Double> result = new ArrayList<>();
        Deque<TreeNode> deque = new LinkedList<>();

        if(root == null) {
            return result;
        }
        deque.offer(root);

        while(!deque.isEmpty()) {
            int size = deque.size();
            double sum = 0;
            for(int i = 0; i < size; i++) {
                TreeNode cur = deque.poll();
                sum += cur.val;
                if(cur.left != null) {
                    deque.offer(cur.left);
                }
                if(cur.right != null) {
                    deque.offer(cur.right);
                }
            }
            result.add(sum / size);
        }
        return result;
    }
}

429.N叉树的层序遍历

题目链接：N叉树的层序遍历

二叉树层序遍历，左右结点放进deque
N叉数层序遍历，子节点集合list遍历一个一个放入deque

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> result = new ArrayList<>();
        Deque<Node> deque = new LinkedList<>();

        if(root == null) {
            return result;
        }
        deque.offer(root);

        while(!deque.isEmpty()) {
            int size = deque.size();
            List<Integer> list = new ArrayList<>();
            for(int i = 0; i < size; i++) {
                Node cur = deque.poll();
                list.add(cur.val);
                if(cur.children != null) {
                    for(Node node : cur.children) {
                        deque.add(node);
                    }
                }
            }
            result.add(list);
        }
        return result;
    }
}