Link: https://leetcode.cn/problems/inorder-successor-in-bst/
Question
difficulty: mid
adj diff: 4
Given the root of a binary search tree and a node p in it, return the in-order successor of that node in the BST. If the given node has no in-order successor in the tree, return null.
The successor of a node p is the node with the smallest key greater than p.val.
Example 1:
Input: root = [2,1,3], p = 1
Output: 2
Explanation: 1's in-order successor node is 2. Note that both p and the return value is of TreeNode type.
Example 2:
Input: root = [5,3,6,2,4,null,null,1], p = 6
Output: null
Explanation: There is no in-order successor of the current node, so the answer is null.
Constraints:
The number of nodes in the tree is in the range [1, 104].
-105 <= Node.val <= 105
All Nodes will have unique values.
主要是学会怎么用 Stack 来写 binary tree 的中序遍历。
key:每次都推整个 left branch 入栈。
Code
class Solution {
public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
Stack<TreeNode> stack = new Stack<TreeNode>();
// push the entire left branches
while (root != null) {
stack.push(root);
root = root.left;
}
boolean foundPreNode = false;
while (!stack.isEmpty()) {
TreeNode nextNode = stack.pop();
if (foundPreNode) {
return nextNode;
} else if (nextNode == p) {
foundPreNode = true;
}
// if have right child, push left branches of right child
if (nextNode.right != null) {
stack.push(nextNode.right);
while (stack.peek().left != null) {
stack.push(stack.peek().left);
}
}
}
return null;
}
}