LeetCode题解：Binary Tree Level Order Traversal I and II

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]]

Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

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

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

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

思路：

把栈式的访问改成队列式的访问。深度优先搜索改为广度优先搜索。

题解：

/** * 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) {        if (!root)            return vector<vector<int>>();                queue<pair<TreeNode*, int>> visit_list;        vector<vector<int>> ret;                visit_list.push(make_pair(root, 0));        while(!visit_list.empty())        {            while(visit_list.front().second >= ret.size())                ret.push_back(vector<int>());            ret[visit_list.front().second].push_back(visit_list.front().first->val);                        if (visit_list.front().first->left != nullptr)                visit_list.push(make_pair(visit_list.front().first->left, visit_list.front().second + 1));            if (visit_list.front().first->right != nullptr)                visit_list.push(make_pair(visit_list.front().first->right, visit_list.front().second + 1));            visit_list.pop();        }                return ret;    }};

思路：

同上，只不过多一次std::reverse的工作。注意这里用reverse并不会带来vector的值交换，而只是交换vector内部的指针而已，所以性能没有什么损失。

题解：

/** * 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> > levelOrderBottom(TreeNode *root) {        vector<vector<int>> retval;        if (root == nullptr)            return retval;                queue<pair<TreeNode*, int>> traverse;        traverse.push(make_pair(root, 1));        while(!traverse.empty())        {            auto front = traverse.front();            traverse.pop();                        if (retval.size() < front.second)                retval.push_back(vector<int>());            retval.back().push_back(front.first->val);                        if (front.first->left !=  nullptr)                traverse.push(make_pair(front.first->left, front.second + 1));            if (front.first->right != nullptr)                traverse.push(make_pair(front.first->right, front.second + 1));        }                // since std::swap is used         // there is actually no vector swap, just pointers swap        reverse(begin(retval), end(retval));                return retval;    }};