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BSTIterator173.java
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BSTIterator173.java
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package Leetcode;
import java.util.ArrayList;
import java.util.List;
import java.util.Stack;
public class BSTIterator173 {
/**Method1: ArrayList store by inOrder; Time: Constructor-O(N), next()-O(1), hasNext()-O(1); Space: O(N) **/
/**
* 中序遍历:
1. 直接递归中序遍历
2. 把中序内容存到list中
3. 逐个返回list的值
**/
List<Integer> list;
int index;
public BSTIterator173(TreeNode root) {
this.list = new ArrayList<>();
index = 0;
inOrder(root);
}
private void inOrder(TreeNode root) {
if(root == null) {
return;
}
inOrder(root.left);
this.list.add(root.val);
inOrder(root.right);
}
public int next() {
return this.list.get(index++);
}
public boolean hasNext() {
return index < this.list.size();
}
/**Method2: Store data in an stack; Time: Constructor-O(H), next()-O(H), hasNext()-O(1); Space: O(H) **/
class BSTIterator {
/**
* 中序遍历:
1. 先把根节点的左子节点都入栈: 例如,先把7, 3分别入栈,
2. stack中最上面的节点就是最左节点,就是next的输出节点,最上节点为3,
3. 同时要考虑next如果有右子树,要作为next的下一个元素
**/
Stack<TreeNode> stack;
TreeNode cur;
public BSTIterator(TreeNode root) {
stack = new Stack<TreeNode>();
cur = root;
}
public int next() {
while(cur!=null) {
stack.push(cur);
cur = cur.left;
}
cur = stack.pop();
int res = cur.val;
cur = cur.right;
return res;
}
public boolean hasNext() {
return !stack.isEmpty() || cur != null;
}
}
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {}
TreeNode(int val) { this.val = val; }
TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}
}