LeetCode（剑指 Offer）- 32 - II. 从上到下打印二叉树 II

题目链接：​​点击打开链接​​

题目大意：略

解题思路：略

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AC 代码

• Java

/**
 * 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;
 *     }
 * }
 */
 
// 解决方案(1)
class Solution {
 
    private List<List<Integer>> list;
 
    private Queue<TreeNode> queue;
 
    private Map<TreeNode, Integer> map;
 
    public List<List<Integer>> levelOrder(TreeNode root) {
        list = new ArrayList<>();
        if (null == root) {
            return list;
        }
        map = new HashMap<TreeNode, Integer>(){{put(root, 0);}};
        queue = new LinkedList<TreeNode>(){{offer(root);}};
        dfs(root, 0);
        bfs();
        return list;
    }
 
    private void bfs() {
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            Integer index = map.get(node);
            if (list.size() <= index) {
                list.add(new ArrayList<>());
            }
            list.get(index).add(node.val);
 
            if (null != node.left) {
                queue.offer(node.left);
            }
            if (null != node.right) {
                queue.offer(node.right);
            }
        }
    }
 
    private void dfs(TreeNode node, int index) {
        map.put(node, index);
        if (null != node.left) {
            dfs(node.left, index + 1);
        }
        if (null != node.right) {
            dfs(node.right, index + 1);
        }
    }
}
 
// 解决方案(2)
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        List<List<Integer>> res = new ArrayList<>();
        if(root != null) queue.add(root);
        while(!queue.isEmpty()) {
            List<Integer> tmp = new ArrayList<>();
            for(int i = queue.size(); i > 0; i--) {
                TreeNode node = queue.poll();
                tmp.add(node.val);
                if(node.left != null) queue.add(node.left);
                if(node.right != null) queue.add(node.right);
            }
            res.add(tmp);
        }
        return res;
    }
}

• C++

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        queue<TreeNode*> que;
        vector<vector<int>> res;
        int cnt = 0;
        if(root != NULL) que.push(root);
        while(!que.empty()) {
            vector<int> tmp;
            for(int i = que.size(); i > 0; --i) {
                root = que.front();
                que.pop();
                tmp.push_back(root->val);
                if(root->left != NULL) que.push(root->left);
                if(root->right != NULL) que.push(root->right);
            }
            res.push_back(tmp);
        }
        return res;
    }
};