首页 > 代码库 > LeetCode Binary Tree Level Order Traversal

LeetCode Binary Tree Level Order Traversal

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]]
/** * Definition for binary tree * 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> > result;        result.clear();        if(root == NULL)            return result;                vector<TreeNode *> q[2];        int current = 1;        int next = 0;        q[0].push_back(root);                while(q[next].empty() == false)        {            current = !current;            next = !next;            vector<int> tempVec;            for(auto w : q[current])            {                tempVec.push_back(w->val);                if(w->left)                {                    q[next].push_back(w->left);                }                if(w->right)                {                    q[next].push_back(w->right);                }            }            result.push_back(tempVec);            tempVec.clear();            q[current].clear();        }                                return result;    }};

解题思路:层次遍历,在处理层次遍历的代码中,巧妙的运用2个向量保存当前正在处理的层的节点q[current],和该层节点的儿子所组成的下一层节点q[next]在处理开始将current和next互换,轻松实现层次切换。再有就是遍历向量的写法for(auto w :q[current])是c++11的新特性,在编译器中不用通过。

LeetCode Binary Tree Level Order Traversal