May LeetCoding Challenge20 之 二叉树中序遍历

因为题目为二叉搜索树找到第k大的数，所以对二叉搜索树中序遍历一定是有序的。

分为两种：递归 和 迭代(栈)

JAVA

/**
 * Definition for a binary tree node.
 * 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;
 *     }
 * }
 */
class Solution {
    public LinkedList<Integer> inorder(TreeNode root, LinkedList<Integer> arr){
        if(root == null) return arr;
        inorder(root.left, arr);
        arr.add(root.val);
        inorder(root.right, arr);
        return arr;
    }
    public int kthSmallest(TreeNode root, int k) {
        LinkedList<Integer> nums = inorder(root, new LinkedList<>());
        return nums.get(k-1);
    }
}

class Solution {
    public int kthSmallest(TreeNode root, int k) {
        Deque<TreeNode> stack = new LinkedList<>();
        List<Integer> nums = new LinkedList<>();
        TreeNode cur = root;
        while(!stack.isEmpty() || cur != null){
            while(cur != null){
                stack.addLast(cur);
                cur = cur.left;
            }
            cur = stack.removeLast();
            nums.add(cur.val);
            cur = cur.right;
        }
        return nums.get(k-1);
    }
}

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def kthSmallest(self, root: TreeNode, k: int) -> int:
        num = []
        return self.inorder(root, num)[k-1]
    def inorder(self, root, res):
        if root == None:
            return res
        self.inorder(root.left, res)
        res.append(root.val)
        self.inorder(root.right, res)
        return res

class Solution:
    def kthSmallest(self, root: TreeNode, k: int) -> int:
        stack = []
        res = []
        cur = root
        while stack or cur:
            if cur:
                stack.append(cur)
                cur = cur.left
            else: 
                cur = stack.pop()
                res.append(cur.val)
                cur = cur.right
        return res[k-1]