LeetCode -- 二叉树水题II

Binary Tree Level Order Traversal

链接：http://leetcode.com/onlinejudge#question_102

原题：

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：按照一层层遍历就可以了。

代码：

/** * 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<vector<int> > collection;        if (root == NULL)            return collection;        vector<int> vec;        vector<vector<TreeNode*> > layers;        vector<TreeNode*> layer;        layer.push_back(root);        layers.push_back(layer);                while (true) {            vector<TreeNode*> &preLayer = layers.back();            vec.clear();            layer.clear();            for (int i=0; i<preLayer.size(); i++) {                if (preLayer[i]) {                    vec.push_back(preLayer[i]->val);                    layer.push_back(preLayer[i]->left);                    layer.push_back(preLayer[i]->right);                }            }            if (vec.size() == 0)                break;            collection.push_back(vec);            layers.push_back(layer);        }        return collection;    }};

 Unique Binary Search Trees

链接：http://leetcode.com/onlinejudge#question_96

原题：

Given n, how many structurally unique BST's (binary search trees) that store values 1...n?

For example,
Given n = 3, there are a total of 5 unique BST's.

   1         3     3      2      1    \       /     /      / \      \     3     2     1      1   3      2    /     /       \                 \   2     1         2                 3

 

思路：只要求出递推式就可以了。

对于n个数字，记树的个数为treeNum[n];

树根只有n种可能，如果树根为i, 那么左子树有i-1个结点, 右子树有n-i个结点,

一共有treeNum[i-1] * treeNum[n-i]棵树, i = 1, ...., n, 累加一下就可以了.

代码:

class Solution {public:    int numTrees(int n) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (n == 0)            return 0;        vector<int> counter(n+1);        counter[0] = 1;        counter[1] = 1;        for (int k=2; k<=n; k++) {            int total = 0;            for (int i=1; i<=k; i++)                total += counter[i-1]*counter[k-i];            counter[k] = total;        }                return counter[n];    }};

Populating Next Right Pointers in Each Node

链接：http://leetcode.com/onlinejudge#question_116

原题：

Given a binary tree

    struct TreeLinkNode {      TreeLinkNode *left;      TreeLinkNode *right;      TreeLinkNode *next;    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

         1       /  \      2    3     / \  / \    4  5  6  7

After calling your function, the tree should look like:

         1 -> NULL       /  \      2 -> 3 -> NULL     / \  / \    4->5->6->7 -> NULL

思路：因为是完全二叉树，所以简单多了。直接父亲节点往next遍历，把相应的子节点连接起来。

代码：

/** * Definition for binary tree with next pointer. * struct TreeLinkNode { *  int val; *  TreeLinkNode *left, *right, *next; *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} * }; */class Solution {public:    void connect(TreeLinkNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (root == NULL || root->left == NULL)    return;TreeLinkNode *head = root->left;TreeLinkNode *parent = root;head->next = parent->right;TreeLinkNode *child = head->next;while (parent->next) {parent = parent->next;child->next = parent->left;parent->left->next = parent->right;child = parent->right;}connect(head);    }};

Populating Next Right Pointers in Each Node II

链接：http://leetcode.com/onlinejudge#question_117

原题：上一道题目的扩展版本，即对任意的二叉树进行connection

思路：其实是一回事情，父亲这一层按照next遍历，随带把子节点connect起来，再递归的做已经连起来的

子节点。

代码：

/** * Definition for binary tree with next pointer. * struct TreeLinkNode { *  int val; *  TreeLinkNode *left, *right, *next; *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} * }; */class Solution {public:    void connect(TreeLinkNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (root == NULL)    return;TreeLinkNode *head = getFirstChild(root);if (head == NULL)return;TreeLinkNode *child = head;if (root->right && child==root->left) {child->next = root->right;child = child->next;}while (root->next) {root = root->next;if (root->left) {child->next = root->left;child = child->next;}if (root->right) {child->next = root->right;child = child->next;}}connect(head);    }private:TreeLinkNode* getFirstChild(TreeLinkNode *&root) {while (root) {if (root->left)return root->left;if (root->right)return root->right;root = root->next;}return NULL;}};