题目
在二叉查找树中插入节点
给定一棵二叉查找树和一个新的树节点,将节点插入到树中。
你需要保证该树仍然是一棵二叉查找树。
样例
给出如下一棵二叉查找树,在插入节点6之后这棵二叉查找树可以是这样的:
2 2
/ \ / \
1 4 --> 1 4
/ / \
3 3 6
分析
递归和非递归两种方法实现。
Python代码
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
"""
@param root: The root of the binary search tree.
@param node: insert this node into the binary search tree.
@return: The root of the new binary search tree.
"""
def insertNode1(self, root, node):
# write your code here
if root is None:
root = node
return root
if node.val < root.val:
root.left = self.insertNode1(root.left, node)
else:
root.right = self.insertNode1(root.left, node)
return root
def insertNode(self, root, node):
# write your code here
if root is None:
root = node
return root
t = root
while t is not None:
if node.val < t.val:
if t.left is None:
t.left = node
return root
else:
t = t.left
continue
else:
if t.right is None:
t.right = node
return root
else:
t = t.right
continue
return root
C++代码
/**
* Definition of TreeNode:
* class TreeNode {
* public:
* int val;
* TreeNode *left, *right;
* TreeNode(int val) {
* this->val = val;
* this->left = this->right = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param root: The root of the binary search tree.
* @param node: insert this node into the binary search tree
* @return: The root of the new binary search tree.
*/
TreeNode* insertNode1(TreeNode* root, TreeNode* node) {
// write your code here
if(root == NULL)
{
root = node;
return root;
}//if
if(node->val < root->val)
{
if(root->left == NULL)
{
root->left = node;
}else{
root->left = insertNode1(root->left, node);
}//else
}else{
if(root->right == NULL)
{
root->right = node;
}else{
root->right = insertNode1(root->right, node);
}//else
}//else
return root;
}
//非递归
TreeNode* insertNode(TreeNode* root, TreeNode* node) {
// write your code here
if(root == NULL)
{
root = node;
return root;
}//if
TreeNode *t = root;
while(t != NULL)
{
if(node->val < t->val)
{
if(t->left == NULL)
{
t->left = node;
return root;
}else{
t = t->left;;
continue;
}//else
}//if
else{
if(t->right == NULL)
{
t->right = node;
return root;
}else{
t = t->right;
continue;
}//else
}//else
}//while
return root;
}
};