530. Minimum Absolute Difference in BST

Given a binary search tree with non-negative values, find the minimum absolute difference between values of any two nodes.

Example:

Input:   1    \     3    /   2Output:1Explanation:The minimum absolute difference is 1, which is the difference between 2 and 1 (or between 2 and 3).

Note:There are at least two nodes in this BST.

我的解答，想得有点复杂，当时做的时候想的是，最小的差会有两种情况，一种式节点与其左子树中的最右子树的差值，以及与其右子树中的最左子树的差值。

写出来的比较复杂。

/** * 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 getMinimumDifference(TreeNode* root) {        queue<TreeNode*> q;        q.push(root);        int rs = 0x7fffffff;        while(!q.empty()){            TreeNode* temp = q.front();            q.pop();            rs = (rs < min(minLeftTree(temp), minRightLeft(temp)) ? rs : min(minLeftTree(temp), minRightLeft(temp)));            if(temp->left) q.push(temp->left);            if(temp->right) q.push(temp->right);        }        return rs;            }        //对于每个节点，和他差值最小的是min(左子树的最右子树节点，右子树的最左子树节点)    int minLeftTree(TreeNode* root){        if(!root || !root->left){            return 0x7fffffff;        }        if(!root->left->right){            return root->val - root->left->val;        }        TreeNode* temp = root->left->right;        while(temp->right){            temp = temp->right;        }        return root->val - temp->val;    }        int minRightLeft(TreeNode* root){        if(!root || !root->right){            return 0x7fffffff;        }        if(!root->right->left){            return root->right->val - root->val;        }        TreeNode* temp = root->right->left;        while(temp->left){            temp = temp->left;        }        return temp->val - root->val;    }};

最简单的方式，是用中序遍历，因为其为二叉搜索数，所以中序遍历的结果得到的节点便是一个从小递增的过程，

则此时最小的差值肯定出现在相邻的两个点之见。

/** * 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 getMinimumDifference(TreeNode* root) {        int val = -1, min_diff = INT_MAX;        subResult(root, val, min_diff);        return min_diff;    }        void subResult(TreeNode* root, int& val, int& min_diff){        if(root->left){            subResult(root->left, val, min_diff);        }        if(val >= 0){            min_diff = min(min_diff, root->val - val);        }        val = root->val;        if(root->right){            subResult(root->right, val, min_diff);        }    }};