[LeetCode] 285. Inorder Successor in BST 二叉搜索树中的中序后继节点
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, return null.
给一个二叉搜索树和它的一个节点，找出它的中序后继节点，如果没有返回null。
解法1: 用中序遍历二叉搜索树，当找到root.val = p.val的时候，返回下一个节点。T: O(n)， S: O(n)
解法2: 利用BST的性质，不用遍历全部元素。比较root.val和p.val，然后在左子树或者右子树中查找。T: O(h)， S: O(1)
如果节点p有右子树，那么p的后继节点为右子树的的最左值（即最后一个没有左子树的节点）；
如果节点p没有右子树，就在二叉搜索树中搜索节点p，并在从上往下的查找过程中依次更新记录比p的val值大的节点。
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
if(root == null || p == null) {
return null;
}
boolean foundNodeP = false;
Stack<TreeNode> stack = new Stack<>();
while(root != null || !stack.isEmpty()) {
if(root != null) {
stack.push(root);
root = root.left;
} else {
root = stack.pop();
if(foundNodeP) {
return root;
}
if(root.val == p.val) {
foundNodeP = true;
}
root = root.right;
}
}
return null;
}
}
Java:
public class Solution {
public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
if(root == null || p == null) {
return null;
}
TreeNode successor = null;
while(root != null) {
if(p.val < root.val) {
successor = root;
root = root.left;
} else {
root = root.right;
}
}
return successor;
}
}
Java:
public class Solution {
public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
TreeNode node = root, successor = null;
while (node != null) {
if (node.val > p.val) {
successor = node;
node = node.left;
} else {
node = node.right;
}
}
return successor;
}
}
Python:
class Solution(object):
def inorderSuccessor(self, root, p):
"""
:type root: TreeNode
:type p: TreeNode
:rtype: TreeNode
"""
# If it has right subtree.
if p and p.right:
p = p.right
while p.left:
p = p.left
return p
# Search from root.
successor = None
while root and root != p:
if root.val > p.val:
successor = root
root = root.left
else:
root = root.right
return successor
C++:
class Solution {
public:
TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
TreeNode *res = NULL;
while (root) {
if (root->val > p->val) {
res = root;
root = root->left;
} else root = root->right;
}
return res;
}
};
C++: Recursion
class Solution {
public:
TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
if (!root) return NULL;
if (root->val <= p->val) {
return inorderSuccessor(root->right, p);
} else {
TreeNode *left = inorderSuccessor(root->left, p);
return left ? left : root;
}
}
};
类似题目：
[LeetCode] 94. Binary Tree Inorder Traversal 二叉树的中序遍历
[LeetCode] 173. Binary Search Tree Iterator 二叉搜索树迭代器
