目錄

• • 題目
  • 思路1：遞歸遍歷得到result數(shù)組(單調(diào)遞增)，然后對(duì)數(shù)組進(jìn)行前后差分，取最小值
  • 思路2：不用數(shù)組，進(jìn)行優(yōu)化
  • 思路3、回顧迭代法求解

題目

給你一棵所有節(jié)點(diǎn)為非負(fù)值的二叉搜索樹(shù)，請(qǐng)你計(jì)算樹(shù)中任意兩節(jié)點(diǎn)的差的絕對(duì)值的最小值。

思路1：遞歸遍歷得到result數(shù)組(單調(diào)遞增)，然后對(duì)數(shù)組進(jìn)行前后差分，取最小值

/*** 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 traversal(TreeNode* cur , vector<int>& vec){if(cur == NULL) return;traversal(cur->left,vec);vec.push_back(cur->val);traversal(cur->right,vec);}int getMinimumDifference(TreeNode* root) {vector<int> result;traversal(root,result);//遞增序列，所以絕對(duì)值的最小值就在差分?jǐn)?shù)組中int Minimum =INT_MAX;for(int i =1;i<result.size();i++){Minimum = min(result[i]-result[i-1],Minimum);}return Minimum;} };

不過(guò)這樣的效率感覺(jué)比較低。

思路2：不用數(shù)組，進(jìn)行優(yōu)化

/*** 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 Minimum =INT_MAX; TreeNode* pre =NULL; //用來(lái)記錄前一個(gè)結(jié)點(diǎn)void traversal(TreeNode* cur){if(cur == NULL) return;traversal(cur->left);if(pre != NULL ) Minimum = min(cur->val-pre->val,Minimum);pre = cur; //更新結(jié)點(diǎn)traversal(cur->right);}int getMinimumDifference(TreeNode* root) {traversal(root);return Minimum;} };

思路3、回顧迭代法求解

/*** 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) {stack<TreeNode*> st;int Minimum =INT_MAX; TreeNode* cur = root;TreeNode* pre = NULL; while(!st.empty() || cur!=NULL){if(cur!=NULL){st.push(cur);cur = cur->left;}else{cur = st.top();st.pop();if(pre != NULL ) Minimum = min(cur->val-pre->val,Minimum);pre = cur; //更新結(jié)點(diǎn)cur = cur->right;}} return Minimum;} };

總結(jié)

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