二叉树删除边


https://docs.google.com/document/d/1qxA2wps0IhVRWULulQ55W4SGPMu2AE5MkBB37h8Dr58/edit#heading=h.jb1562yh5s6d

2. 二叉树删除边 (高频 13次)

LC684,BT删除多余边
思路:
3 invalid situations
case1: 2 parents no circle
case2: 2 parents with circle
case3: 1 parent with circle
2 main steps
1 check whether there exists a node with 2 parents, if yes, store the two edges.
2 if no edge yielded from step 1, apply union-find and find the edge creating cycle (same as 684); ELSE, apply union-find to ALL edges EXCEPT edges from step 1, then check: if edge 1 from step 1 creates a cycle, return edge 1; else return edge 2.
//请问有的题是binary tree删多余边,这时候输入是 root 还是 edge list ?

Follow up: 给一棵二叉搜索树,有一条多余边,删除它

例子:
    7
  / \
 5   9
/  \ /  
3    8
对于多余边5-8,9-8此处的删除需要有选择,跟之前的题目找到多余边立马不分选择删除有区别

思路:  LC 98: Validate Binary Search Tree
DFS,参数中带着左右边界,返回值为新树的根节点,如果当前节点不在当前树的范围内,返回null删除该边

参考代码
provider: null
public void deleteEdge(TreeNode root) {
if(root == null) return;
root = dfs(root, Integer.MIN_VALUE, Integer.MAX_VALUE);
}
private TreeNode dfs(TreeNode root, int left, int right) {
if(root == null) return null;
if(root.val <= left || root.val >= right) return null;
root.left = dfs(root.left, left, root.val);
root.right = dfs(root.right, root.val, right);
return root;
}
能否说一句无关的废话,上面BST删除多余的edge的思路,和BST insert以及delete node的思路很类似,可以放到一起总结复习。(Editor: 嗯,思路差不多,但是follow up用DFS感觉好做一点,第一题UF和DFS都能做)

CPP参考代码:
(Provider: Anonym)
void isValidBST(TreeNode* root) {
   root=isValidBST(root, nullptr, nullptr);
}
TreeNode* isValidBST(TreeNode* root, TreeNode* minNode, TreeNode* maxNode) {
if(!root) return nullptr;
if(minNode && root->val <= minNode->val || maxNode && root->val >= maxNode->val)
       return nullptr;
   root->left = isValidBST(root->left, minNode, root);
   root->right = isValidBST(root->right, root, maxNode);
   return root;

}



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