LeetCode 701 Insert into a Binary Search Tree (DFS)

最新推荐文章于 2023-02-25 22:41:55 发布

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本文介绍了一种高效的方法，用于将一个新值插入到二叉搜索树中，同时保持树的搜索特性不变。文章提供了两种解决方案，一种是通过寻找父节点进行插入，另一种是使用递归直接在树中找到正确的位置。最终，文章提供了一个0ms的解决方案，时间性能击败了100%的同类算法。

Given the root node of a binary search tree (BST) and a value to be inserted into the tree, insert the value into the BST. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.

Note that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any of them.

For example,

Given the tree:
        4
       / \
      2   7
     / \
    1   3
And the value to insert: 5

You can return this binary search tree:

This tree is also valid:

题目链接：https://leetcode.com/problems/insert-into-a-binary-search-tree/

题目分析：一开始的做法很麻烦，单独写了个函数找父节点

0ms，时间击败100%

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
    public TreeNode findParent(TreeNode root, int val) {
        if (root.left == null && root.right == null) {
            return root;
        }
        if (root.val < val) {
            if (root.right != null) {
                return findParent(root.right, val);
            } else {
                return root;
            }
        }
        if (root.left != null) {
            return findParent(root.left, val);
        }
        return root;
    }
    
    public TreeNode insertIntoBST(TreeNode root, int val) {
        if (root == null) {
            return new TreeNode(val);
        }
        TreeNode parent = findParent(root, val);
        if (val < parent.val) {
            parent.left = new TreeNode(val);
        } else {
            parent.right = new TreeNode(val);
        }
        return root;
    }
}

其实直接原地递归找就行了

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode insertIntoBST(TreeNode root, int val) {
        if (root == null) {
            return new TreeNode(val);
        }
        if (root.val < val) {
            root.right = insertIntoBST(root.right, val);
        } else {
            root.left = insertIntoBST(root.left, val);
        }
        return root;
    }
}