层次遍历二叉树

标签：

    3
   /   9  20
    /     15   7

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

public class Solution {

  static class Queue {
        static final int MAX_SIZE = 1 << 10;
        private TreeNode list[];
        int head;
        int tail;
        int cap;

        Queue() {
            list = new TreeNode[MAX_SIZE + 1];
            head = 0;
            tail = 0;
            cap = MAX_SIZE + 1;
        }

        Queue(int size) {
            list = new TreeNode[size + 1];
            head = 0;
            tail = 0;
            cap = size + 1;
        }

        boolean isEmpty() {
            return tail == head;
        }

        boolean isFull() {
            return (tail + 1) % cap == head;
        }

        void push(TreeNode node) {
            list[tail] = node;
            tail = (tail + 1) % cap;
        }

        TreeNode pop() {
            TreeNode top = list[head];
            head = (head + 1) % cap;
            return top;
        }

        int size() {
            return (tail - head + cap) % cap;
        }

        TreeNode top() {
            return list[head];

        }

        void print() {
            for (int i = head; i % cap != tail; i = (i + 1) % cap) {
                System.out.print(list[i].val + ",");
            }
            System.out.printf("\n");
        }
    }

    static public boolean judge(LinkedList<List<TreeNode>> lists, Queue queue) {
        if (lists.isEmpty()) return true;
        if (queue.isEmpty()) return false;
        return !lists.getLast().contains(queue.top());
    }

    static public void addQueue2List(Queue queue, List<List<Integer>> lists, LinkedList<List<TreeNode>> nodes) {
        List<Integer> vals = new ArrayList<Integer>();
        List<TreeNode> nos = new ArrayList<TreeNode>();
        for (int i = queue.head; i % queue.cap != queue.tail; i = (i + 1) % queue.cap) {
            vals.add(queue.list[i].val);
            nos.add(queue.list[i]);
        }
        lists.add(vals);
        nodes.add(nos) ;
    }

     public List<List<Integer>> levelOrder(TreeNode root) {
        Queue queue = new Queue();
        LinkedList<List<Integer>> lists = new LinkedList<List<Integer>>();
        LinkedList<List<TreeNode>> nodes = new LinkedList<List<TreeNode>>();

        if (root != null) {
            queue.push(root);
        }


        while (!queue.isEmpty()) {
            //queue.print();
            if (judge(nodes, queue)) {
                addQueue2List(queue, lists, nodes);
            }

            TreeNode node = queue.pop();

            if (node.left != null) {
                queue.push(node.left);
            }

            if (node.right != null) {
                queue.push(node.right);
            }

        }


        return lists;
    }


}

基本思路：观察每次进出队列时候的队列快照，发现在某一时刻，队列中的节点恰好为未访问的整个一层的节点。

核心在judge方法，会去检查已经访问的上一层次的节点中是否包含当前队列中的节点。只有judge返回true，

说明上一层的节点全部都出队了，可以安心访问下一层次的节点了。

层次遍历二叉树

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原文地址：http://blog.csdn.net/wongson/article/details/44982503