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  • [leedcode 173] Binary Search Tree Iterator

    Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

    Calling next() will return the next smallest number in the BST.

    Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

    /**
     * Definition for binary tree
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    
    public class BSTIterator {
        Stack<TreeNode> stack=new Stack<TreeNode>();
        //中序遍历+stack
        //非递归中序遍历二叉树的话就是借助栈,让下次输出的目标节点始终存放在栈顶位置。每次输出一个节点,
        //就将此节点的后续节点放入栈中(沿着右子树的左子树一直向左就是下一个要输出的节点)。
        public BSTIterator(TreeNode root) {
            pushStack(root);
        }
    
        /** @return whether we have a next smallest number */
        public boolean hasNext() {
            return !stack.empty();
        }
    
        /** @return the next smallest number */
        public int next() {
            TreeNode node=stack.pop();
            pushStack(node.right);///注意保存右节点
            return node.val;
        }
        private void pushStack(TreeNode node){
                while(node!=null){
                    stack.push(node);
                    node=node.left;
                }
        } 
    }
    
    /**
     * Your BSTIterator will be called like this:
     * BSTIterator i = new BSTIterator(root);
     * while (i.hasNext()) v[f()] = i.next();
     */
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  • 原文地址:https://www.cnblogs.com/qiaomu/p/4696398.html
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