寻找下一个结点 牛客网 程序员面试金典 C++ java Python
寻找下一个结点 牛客网 程序员面试金典 C++ java Python
- 题目描述
- 请设计一个算法,寻找二叉树中指定结点的下一个结点(即中序遍历的后继)。
- 给定树的根结点指针TreeNode* root和结点的值int p,请返回值为p的结点的后继结点的值。保证结点的值大于等于零小于等于100000且没有重复值,若不存在后继返回-1。
C++
/*struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};*/
class Successor {
//run:5ms memory:504k
TreeNode* pre = new TreeNode(-1);
public:
int findSucc(TreeNode* root, int p){
if (NULL == root) return -1;
int ret = findSucc(root->left,p);
if (-1 == ret){
if (pre->val == p) return root->val;
pre = root;
return findSucc(root->right,p);
}
return ret;
}
};
java
import java.util.*;/*
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}*/
public class Successor {
//run:32ms memory:10444k
private TreeNode pre = new TreeNode(-1);
public int findSucc(TreeNode root, int p) {
if (root == null) return -1;
int ret = findSucc(root.left, p);
if (ret == -1) {
if (pre.val == p) return root.val;
pre = root;
return findSucc(root.right, p);
}
return ret;
}
}
Python
python"># class TreeNode:# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Successor:
#run:43ms memory:5856k
def __init__(self):
self.pre = TreeNode(-1)
def findSucc(self, root, p):
if None == root: return -1
ret = self.findSucc(root.left,p)
if -1 == ret:
if self.pre.val == p: return root.val
self.pre = root
return self.findSucc(root.right,p)
return ret
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