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  • leetcode-173:Binary Search Tree Iterator(Java)

    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.

    Credits:
    Special thanks to @ts for adding this problem and creating all test cases.

    要求:写一个二叉查找树,每次返回树中的下一个最小节点


    比如上图中的二叉查找树,从根节点开始,依次返回1,3,4,6,7... ...

    思路:维护一个栈,先将根结点的左子树全部压栈,每次弹出栈顶元素,若某次弹出的栈顶元素有右子树,比如3,此时需要将以该节点的右子树为根的子树的左子节点全部压栈


    /**
     * Definition for binary tree
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    
    import java.util.Stack;
    public class BSTIterator {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        
        public BSTIterator(TreeNode root) {
            
            while(root != null){
                stack.push(root);
                root = root.left;
            }
        }
    
        /** @return whether we have a next smallest number */
        public boolean hasNext() {
            return !stack.isEmpty();
            
        }
    
        /** @return the next smallest number */
        public int next() {
            TreeNode minCurrent = stack.pop();
            if(minCurrent.right != null){
                TreeNode rightNode = minCurrent.right;
                while(rightNode != null){
                    stack.push(rightNode);
                    rightNode = rightNode.left;
                }
            }
            
           return minCurrent.val; 
        }
    }
    
    /**
     * 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/wennian/p/5036883.html
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