寻找下一个结点 牛客网 程序员面试金典 C++ java Python

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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