[LeetCode]230. Kth Smallest Element in a BST
2017-04-16 22:29
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Given a binary search tree, write a function
smallest element in it.
Note:
You may assume k is always valid, 1 ? k ? BST's total elements.
思路:从最小的开始进行从1编号,直到等于k停止并返回值。
使用一个栈,先从root一直选择左节点到最后为止,都压入栈中,此时栈顶就为最小的节点,表为1,随后开始弹栈
弹栈后先进行判断,如果有右节点,则说明在此节点和此时栈顶之前还有介于这两个值的节点没有遍历,然后把这个右节点的选择左节点到最后压入栈中,最后再将编号加一,进行下一个节点递归
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int kthSmallest(TreeNode root, int k) {
Stack<TreeNode> stack=new Stack<TreeNode>();
TreeNode h=root;
while(h!=null){
stack.push(h);
h=h.left;
}
return fun(stack.pop(),1,k,stack);
}
public int fun(TreeNode root,int n,int k,Stack<TreeNode> stack){
if(n==k){
return root.val;
}
if(root.right!=null){
TreeNode h=root.right;
while(h!=null){
stack.push(h);
h=h.left;
}
}
return fun(stack.pop(),n+1,k,stack);
}
}
kthSmallestto find the kth
smallest element in it.
Note:
You may assume k is always valid, 1 ? k ? BST's total elements.
思路:从最小的开始进行从1编号,直到等于k停止并返回值。
使用一个栈,先从root一直选择左节点到最后为止,都压入栈中,此时栈顶就为最小的节点,表为1,随后开始弹栈
弹栈后先进行判断,如果有右节点,则说明在此节点和此时栈顶之前还有介于这两个值的节点没有遍历,然后把这个右节点的选择左节点到最后压入栈中,最后再将编号加一,进行下一个节点递归
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int kthSmallest(TreeNode root, int k) {
Stack<TreeNode> stack=new Stack<TreeNode>();
TreeNode h=root;
while(h!=null){
stack.push(h);
h=h.left;
}
return fun(stack.pop(),1,k,stack);
}
public int fun(TreeNode root,int n,int k,Stack<TreeNode> stack){
if(n==k){
return root.val;
}
if(root.right!=null){
TreeNode h=root.right;
while(h!=null){
stack.push(h);
h=h.left;
}
}
return fun(stack.pop(),n+1,k,stack);
}
}
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