Leetcode: Binary Tree Level Order Traversal

来源：互联网发布：linux安装oracle9i 编辑：程序博客网时间：2024/06/01 16:15

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]]

两个队列可以解决。

/** * 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;        queue<TreeNode*> first, second;        first.push(root);        TreeNode *cur = NULL;        vector<int> v;        while (!first.empty()) {            while (!first.empty()) {                cur = first.front();                first.pop();                if (cur != NULL) {                    v.push_back(cur->val);                    second.push(cur->left);                    second.push(cur->right);                }            }            if (!v.empty()) {                result.push_back(v);                v.clear();            }                        while (!second.empty()) {                cur = second.front();                second.pop();                if (cur != NULL) {                    v.push_back(cur->val);                    first.push(cur->left);                    first.push(cur->right);                }            }            if (!v.empty()) {                result.push_back(v);                v.clear();            }        }                return result;    }};
=========================第二次==================

一个队列，用NULL来分割。

/** * 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;        queue<TreeNode*> nodes;        vector<int> level;        TreeNode *cur = NULL;                // Use NULL as level seperator        nodes.push(root);        nodes.push(NULL);        while (!nodes.empty()) {            cur = nodes.front();            nodes.pop();            if (cur != NULL) {                if (cur->left != NULL) {                    nodes.push(cur->left);                }                if (cur->right != NULL) {                    nodes.push(cur->right);                }                level.push_back(cur->val);            }            else {                if (!level.empty()) {                    result.push_back(level);                    level.clear();                    nodes.push(NULL);                }            }        }                return result;    }};

0 0