Binary Tree Level Order Traversal -- leetcode

Given a binary tree, return the level order traversal of its nodes‘ values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   /   9  20
    /     15   7

return its level order traversal as:

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

confused what "{1,#,2,3}" means? > rea

使用队列，作FIFO。

在每一层的末尾，插入一个空指针，作为该层结束符。

在leetcode上实际执行时间为11ms。

/**
 * 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) {
        vector<vector<int> > ans;
        if (!root) return ans;
        queue<TreeNode *> q;
        q.push(root);
        q.push(0);
        ans.push_back(vector<int> ());
        while (!q.empty()) {
            root = q.front();
            q.pop();
            if (!root) {
                ans.push_back(vector<int> ());
                if (q.empty()) 
                    break;
                q.push(0);
            }
            else {
                ans.back().push_back(root->val);
                if (root->left) q.push(root->left);
                if (root->right) q.push(root->right);
            }
        }
        ans.pop_back();
        return ans;
    }
};

算法二，深度优先递归遍历

在leetcode上实行际时间为13ms。 比上面的算法慢2ms。但是代码简洁一些。

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int> >ans;
        helper(ans, root, 0);
        return ans;
    }
    
    void helper(vector<vector<int> > &ans, TreeNode *root, int level) {
        if (!root) return;
        if (level == ans.size())
            ans.push_back(vector<int> ());
        
        ans[level].push_back(root->val);
        helper(ans, root->left, level+1);
        helper(ans, root->right, level+1);
    }
};

Binary Tree Level Order Traversal -- leetcode

原文：http://blog.csdn.net/elton_xiao/article/details/45506863