285 Inorder Successor in BST
285. Inorder Successor in BST
1. Question
Given a binary search tree and a node in it, find the in-order successor of that node in the BST.
Note: If the given node has no in-order successor in the tree, returnnull
.
2. Implementation
(1) Inorder traversal
class Solution {
public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
if (root == null) {
return null;
}
Stack<TreeNode> stack = new Stack<>();
TreeNode curNode = root, preNode = null;
while (curNode != null || !stack.isEmpty()) {
if (curNode != null) {
stack.push(curNode);
curNode = curNode.left;
}
else {
curNode = stack.pop();
if (preNode != null && preNode == p) {
return curNode;
}
preNode = curNode;
curNode = curNode.right;
}
}
return null;
}
}
(2) Binary Search
思路: 利用Binary Search Tree的二分特性
class Solution {
public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
if (root == null) {
return null;
}
TreeNode curNode = root;
TreeNode successor = null;
while (curNode != null) {
if (curNode.val <= p.val) {
curNode = curNode.right;
}
else {
successor = curNode;
curNode = curNode.left;
}
}
return successor;
}
}
3. Time & Space Complexity
Inorder Traversal: 时间复杂度: O(n), 空间复杂度: O(h)
Binary Search: 时间复杂度: O(logn), 空间复杂度: O(1)
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