二叉树的层次遍历

给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。

例如:
给定二叉树: [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

返回其层次遍历结果：

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

 

递归法：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levels;
    
    vector<vector<int>> levelOrder(TreeNode* root) {
        helper(root, 0);
        return levels;
    }

    
    void helper(TreeNode* node, int level){
        if(node == NULL) return;
        if(level == levels.size()){
            levels.resize(level+1);
        }
        levels[level].emplace_back(node->val);
        helper(node->left,level+1);
        helper(node->right,level+1);
    }
};

 

迭代法：

 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        queue<TreeNode*> q;
        vector<vector<int>> levels;
        if(root == NULL) return levels;
        q.push(root);
        
        while(!q.empty()){
            int num = q.size();
            vector<int> level;
            for(int i = 0; i < num; i++){
                TreeNode *node = q.front();
                level.emplace_back(node->val);
                q.pop();
                
                if(node->left != NULL) q.push(node->left);
                if(node->right != NULL) q.push(node->right);
            }
            levels.emplace_back(level);
        }
        return levels;
    }
};