LeetCode:Binary Tree Level Order Traversal

题目描写叙述:

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]
]


思路:利用队列层序遍历二叉树。在队列中增加一个flag指针。每次遍历到flag指针时就表示当前这一层遍历结束。


代码:

vector<vector<int> > Solution::levelOrder(TreeNode *root)
{
    vector<vector<int> > result;
    vector<int> sequence;
    TreeNode * flag = new TreeNode(0);
    queue<TreeNode *> treenode_queue;
    if(root == NULL)
        return result;
    treenode_queue.push(root);
    treenode_queue.push(flag);
    while(treenode_queue.size() >= 1)
    {
        TreeNode * node = treenode_queue.front();
        treenode_queue.pop();
        if(node == flag)
        {
            result.push_back(sequence);
            sequence.clear();
            if(treenode_queue.empty())
                break;
            treenode_queue.push(flag);
        }
        else
        {
            sequence.push_back(node->val);
            if(node->left != NULL)
                treenode_queue.push(node->left);
            if(node->right != NULL)
                treenode_queue.push(node->right);
        }
    }
    return result;
}


原文地址:https://www.cnblogs.com/lxjshuju/p/7078146.html