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力扣剑指offer第18天 树

宁静的猫 2022-02-03 阅读 60

53)二叉搜索树中的中序后继

class Solution {
public:
    TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
        TreeNode* res = nullptr;
        TreeNode* cur = root;
        while(cur){
          if(cur->val > p->val){
            res = cur;
            cur = cur->left;
          }else{
            cur = cur->right;
          }
        }
        return res;
    }
};

54)所有大于等于节点的值之和

class Solution{
public:
    TreeNode* convertBST(TreeNode* root){
        int sum=0;
        dfs(root, sum);
        return root;
    }
    void dfs(TreeNode* node, int& sum){
        if(!node) return;
        dfs(node->right, sum);
        sum += node->val;
        node->val = sum;
        dfs(node->left, sum);
    }
};

55)二叉搜索树迭代器

class BSTIterator {
private:
    stack<TreeNode*> stk;
    TreeNode* cur;
public:
    BSTIterator(TreeNode* root) {
      cur = root;
    }
    
    int next() {
      while(cur!=nullptr){
        stk.push(cur);
        cur = cur->left;
      }
      cur = stk.top(); stk.pop();
      int ret = cur->val;
      cur = cur->right;
      return ret;
    }
    
    bool hasNext() {
      return cur!=nullptr || !stk.empty();
    }
};
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