LeetCode 272 Closest Binary Tree Traversal II 解題思路

Youngdze 發(fā)布于2019-08-16 13:26 / 2257人閱讀

摘要：原題網(wǎng)址題意在二叉搜索樹(shù)當(dāng)中找到離最近的個(gè)數(shù)。解題思路由于二叉搜索數(shù)的中序遍歷是有序的，比如例子中的樹(shù)，中序遍歷為。

原題網(wǎng)址：https://leetcode.com/problems...

Given a non-empty binary search tree and a target value, find?k?values in the BST that are closest to the target.

Note:

Given target value is a floating point.
You may assume?k?is always valid, that is:?k?≤ total nodes.
You are guaranteed to have only one unique set of?k?values in the BST that are closest to the target.
Follow up:
Assume that the BST is balanced, could you solve it in less than?O(n) runtime (where?n?= total nodes)?

Hint:

Consider implement these two helper functions:

getPredecessor(N), which returns the next smaller node to N.
getSuccessor(N), which returns the next larger node to N.

Try to assume that each node has a parent pointer, it makes the problem much easier.

Without parent pointer we just need to keep track of the path from the root to the current node using a stack.

You would need two stacks to track the path in finding predecessor and successor node separately.

題意：在二叉搜索樹(shù)當(dāng)中找到離target最近的K個(gè)數(shù)。

解題思路：
由于二叉搜索數(shù)的inorder中序遍歷是有序的，比如例子中的樹(shù)，中序遍歷為[1,2,3,4,5]。我們可以利用這一特性，初始化一個(gè)雙端隊(duì)列Deque，用來(lái)存放k個(gè)數(shù)，然后用遞歸的方式，先走到the most left（也就是例子中的1），不斷的向Deque中加入元素，直到元素裝滿，也就是Deque的size()到k個(gè)了，將當(dāng)前元素與target的距離和隊(duì)列頭部與target的距離進(jìn)行對(duì)比，如果當(dāng)前元素的距離更小，則用Deque的pollFirst()方法將頭部吐出，把當(dāng)前元素從addLast()加入。

Example:

Input: root = [4,2,5,1,3], target = 3.714286, and k = 2

    4
   / 
  2   5
 / 
1   3

Output: [4,3]

代碼如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    /**
     *直接在中序遍歷的過(guò)程中完成比較，當(dāng)遍歷到一個(gè)節(jié)點(diǎn)時(shí)，
     如果此時(shí)結(jié)果數(shù)組不到k個(gè)，我們直接將此節(jié)點(diǎn)值加入res中，
     如果該節(jié)點(diǎn)值和目標(biāo)值的差值的絕對(duì)值小于res的首元素和目標(biāo)值差值的絕對(duì)值，
     說(shuō)明當(dāng)前值更靠近目標(biāo)值，則將首元素刪除，末尾加上當(dāng)前節(jié)點(diǎn)值，
     反之的話說(shuō)明當(dāng)前值比res中所有的值都更偏離目標(biāo)值，
     由于中序遍歷的特性，之后的值會(huì)更加的遍歷，所以此時(shí)直接返回最終結(jié)果即可，
     */
    public List closestKValues(TreeNode root, double target, int k) {
        Deque deque = new ArrayDeque<>();
        inorder(root, target, k, deque);
        List res = new ArrayList<>(deque);
        return res;
    }
    private void inorder(TreeNode root, 
                         double target, 
                         int k, 
                         Deque deque) {
        if (root == null) return;
        inorder(root.left, target, k, deque);
        if (deque.size() < k) {
            deque.offer(root.val);
        } else if (Math.abs(root.val - target) < Math.abs(deque.peekFirst()-target) ) {
            deque.pollFirst();
            deque.addLast(root.val);
        } 
        inorder(root.right, target, k, deque);
    }
}

還有一種用Stack完成的方式，思路和遞歸相同，但是iterative的寫(xiě)法，也有必要掌握，必須把控Stack是否為空的情況，當(dāng)前的node為null，但是stack中仍然有元素，依然需要進(jìn)行比較。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List closestKValues(TreeNode root, double target, int k) {
        Deque deque = new ArrayDeque<>();
        Stack stack = new Stack<>();
        TreeNode cur = root;
        while(cur != null || !stack.isEmpty()) {
            while(cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            if (deque.size() < k) {
                deque.addLast(cur.val);
            } else if (Math.abs(cur.val - target) < Math.abs(deque.peekFirst() - target)) {
                deque.pollFirst();
                deque.addLast(cur.val);
            }
            cur = cur.right;
        }
        List res = new ArrayList<>(deque);
        return res;
    }
    
}