算法设计与应用基础-第五周

Diameter of Binary Tree

Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of thelongest
path between any two nodes in a tree. This path may or may not pass through the root.

Example:
Given a binary tree

1
/ \
2   3
/ \
4   5

Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].

Definition: The diameter of a tree (sometimes called the width) is the number of nodes on the longest path between two leaves in the
tree.

The diameter of a tree T is the largest of the following quantities:
the diameter of T's left subtree
the diameter of T's right subtree
the longest path between leaves that goes through the root of T (this can be computed from the heights of the subtrees of T)
根据上面的分析，我们需要求出以树的根结点为根结点的左右子树的高度之和之后，与以树的根结点的左（右）孩子为根结点的左右子树的高度之和进行比较当前这只是一次比较的节点。这个值还需要与树的根结点的左孩子的后代，以及右孩子的后代的不同叶子间的最大距离之和进行比较。
        因此，对树求出直径就是一个不断递归，不断比较的过程。

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int maxdep=0;
int height(const TreeNode* root)
{
if(root==NULL)
return 0;
int left=0;
int right=0;
if(root->left!=NULL)
left=height(root->left)+1;
if(root->right!=NULL)
right=height(root->right)+1;
maxdep=max(maxdep,left+right);
return max(right,left);
}
int diameterOfBinaryTree(TreeNode* root)
{
if(root==NULL) return 0;
else
{
height(root);
return maxdep;
}
}
};

Convert BST to Greater Tree          

Given a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST.

Example:

Input: The root of a Binary Search Tree like this:
5
/   \
2     13

Output: The root of a Greater Tree like this:
18
/   \
20     13

根据题意，我们要找到比该节点大的所有节点再全部相加而获得该节点的新值，由二分查找树的性质，可以采用后序遍历的方式从大到小的获取节点并在此过程中累加到root上（root在递归中不断更新）。

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void LastOrderVisit(TreeNode *root,int &sum)
{
if(root!=NULL)
{
LastOrderVisit(root->right,sum);//右子树
root->val += sum;
sum = root->val;
LastOrderVisit(root->left,sum);//左子树
}

}
TreeNode* convertBST(TreeNode* root)
{
if(root==NULL) return NULL;
if(root)
{
int res=0;
LastOrderVisit(root,res);
return root;
}
}
};