[Algorithm] BST问题

BST性质：左边都比root小，右边都比root大

BST的中序遍历得到的节点访问顺序是从小到大的顺序

98. Valid Binary Search Tree： 点击打开链接

方法一

思路：先中序遍历BST拿到结果list，遍历list里元素，看当前的元素值是不是小于下一个元素值，只要有当前元素值大于等于下一个元素值的，就返回false

/** * Definition of TreeNode: * public class TreeNode { *     public int val; *     public TreeNode left, right; *     public TreeNode(int val) { *         this.val = val; *         this.left = this.right = null; *     } * } */public class Solution {    /**     * @param root: The root of binary tree.     * @return: True if the binary tree is BST, or false     */    public boolean isValidBST(TreeNode root) {        if(root==null){            return true;        }                List<TreeNode> result=inOrder(root);        for(int i=0;i<result.size()-1;i++){            if(result.get(i).val>=result.get(i+1).val){                return false;            }        }        return true;    }        private List<TreeNode> inOrder(TreeNode root){        List<TreeNode> list=new ArrayList<>();                Stack<TreeNode> stack=new Stack<>();        TreeNode cur=root;        while(cur!=null || !stack.isEmpty()){            while(cur!=null){                stack.push(cur);                cur=cur.left;            }            cur=stack.pop();            list.add(cur);            cur=cur.right;        }        return list;    }}

方法二

思路：递归，边遍历边比较

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean isValidBST(TreeNode root) {        return helper(root,Long.MAX_VALUE, Long.MIN_VALUE);            }    public boolean helper(TreeNode root,long maxVal,long minVal){        if(root==null){            return true;        }                if(root.val>=maxVal || root.val<=minVal){            return false;        }        return helper(root.left,root.val,minVal) && helper(root.right,maxVal,root.val);    }  }

11. Search Range in Binary Search Tree:  点击打开链接

思路：中序遍历得到list，再遍历拿到list里每一个符合条件的值

/** * Definition of TreeNode: * public class TreeNode { *     public int val; *     public TreeNode left, right; *     public TreeNode(int val) { *         this.val = val; *         this.left = this.right = null; *     } * } */public class Solution {    /**     * @param root: The root of the binary search tree.     * @param k1 and k2: range k1 to k2.     * @return: Return all keys that k1<=key<=k2 in ascending order.     */    public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {        ArrayList<Integer> result=new ArrayList<>();        if(root==null){            return result;        }                ArrayList<Integer> list=inorder(root);        for(int i=0;i<list.size();i++){            if(list.get(i)>=k1 && list.get(i)<=k2){                result.add(list.get(i));            }        }        return result;    }        private ArrayList<Integer> inorder(TreeNode root){        ArrayList<Integer> result=new ArrayList<>();        if(root==null){            return result;        }        Stack<TreeNode> stack=new Stack<>();                TreeNode cur=root;        while(!stack.isEmpty() || cur!=null){            while(cur!=null){                stack.push(cur);                cur=cur.left;            }            cur=stack.pop();            result.add(cur.val);            cur=cur.right;        }        return result;    }}

173.Binary Search Tree Iterator： 点击打开链接

思路：就是BST中序遍历的拆解形式

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class BSTIterator {    Stack<TreeNode> stack;        public BSTIterator(TreeNode root) {        stack=new Stack<>();        AddNodeToStack(root);    }    /** @return whether we have a next smallest number */    public boolean hasNext() {        return !stack.isEmpty();    }    /** @return the next smallest number */    public int next() {        TreeNode cur=stack.pop();        AddNodeToStack(cur.right);        return cur.val;    }         private void AddNodeToStack(TreeNode root){        while(root!=null){            stack.push(root);            root=root.left;        }    }}/** * Your BSTIterator will be called like this: * BSTIterator i = new BSTIterator(root); * while (i.hasNext()) v[f()] = i.next(); */