Question

Source

• lintcode: (85) Insert Node in a Binary Search Tree

Given a binary search tree  and a new tree node, insert the node into the tree. You should keep the tree still be a valid binary search tree.

Example
Given binary search tree as follow:

   2

  /  \

1    4

     /

     3

after Insert node 6, the tree should be:

   2

  /  \

1    4

     /   \

     3    6

Challenge
Do it without recursion

题解 - 递归

二叉树的题使用递归自然是最好理解的，代码也简洁易懂，缺点就是递归调用时栈空间容易溢出，故实际实现中一般使用迭代替代递归，性能更佳嘛。不过迭代的缺点就是代码量稍（很) 大，逻辑也可能不是那么好懂。

既然确定使用递归，那么接下来就应该考虑具体的实现问题了。在递归的具体实现中，主要考虑如下两点：

1. 基本条件/终止条件 - 返回值需斟酌。
2. 递归步/条件递归 - 能使原始问题收敛。

首先来找找递归步，根据二叉查找树的定义，若插入节点的值若大于当前节点的值，则继续与当前节点的右子树的值进行比较；反之则继续与当前节点的左子树的值进行比较。题目的要求是返回最终二叉搜索树的根节点，从以上递归步的描述中似乎还难以对应到实际代码，这时不妨分析下终止条件。

有了递归步，终止条件也就水到渠成了，若当前节点为空时，即返回结果。问题是——返回什么结果？当前节点为空时，说明应该将「插入节点」插入到上一个遍历节点的左子节点或右子节点。对应到程序代码中即为 root->right = node 或者 root->left = node . 也就是说递归步使用 root->right/left = func(...) 即可。

C++ Recursion

/**
 * forked from http://www.jiuzhang.com/solutions/insert-node-in-binary-search-tree/
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *   int val;
 *   TreeNode *left, *right;
 *   TreeNode(int val) {
 *     this->val = val;
 *     this->left = this->right = NULL;
 *   }
 * }
 */
class Solution {
public:
  /**
   * @param root: The root of the binary search tree.
   * @param node: insert this node into the binary search tree
   * @return: The root of the new binary search tree.
   */
  TreeNode* insertNode(TreeNode* root, TreeNode* node) {
    if (NULL == root) {
      return node;
    }

    if (node->val <= root->val) {
      root->left = insertNode(root->left, node);
    } else {
      root->right = insertNode(root->right, node);
    }

    return root;
  }
};

Java Recursion

public class Solution {
  /**
   * @param root: The root of the binary search tree.
   * @param node: insert this node into the binary search tree
   * @return: The root of the new binary search tree.
   */
  public TreeNode insertNode(TreeNode root, TreeNode node) {
    if (root == null) {
      return node;
    }
    if (root.val > node.val) {
      root.left = insertNode(root.left, node);
    } else {
      root.right = insertNode(root.right, node);
    }
    return root;
  }
}

题解 - 迭代

看过了以上递归版的题解，对于这个题来说，将递归转化为迭代的思路也是非常清晰易懂的。迭代比较当前节点的值和插入节点的值，到了二叉树的最后一层时选择是链接至左子结点还是右子节点。

C++

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *   int val;
 *   TreeNode *left, *right;
 *   TreeNode(int val) {
 *     this->val = val;
 *     this->left = this->right = NULL;
 *   }
 * }
 */
class Solution {
public:
  /**
   * @param root: The root of the binary search tree.
   * @param node: insert this node into the binary search tree
   * @return: The root of the new binary search tree.
   */
  TreeNode* insertNode(TreeNode* root, TreeNode* node) {
    if (NULL == root) {
      return node;
    }

    TreeNode* tempNode = root;
    while (NULL != tempNode) {
      if (node->val <= tempNode->val) {
        if (NULL == tempNode->left) {
          tempNode->left = node;
          return root;
        }
        tempNode = tempNode->left;
      } else {
        if (NULL == tempNode->right) {
          tempNode->right = node;
          return root;
        }
        tempNode = tempNode->right;
      }
    }

    return root;
  }
};

源码分析

在 NULL == tempNode->right 或者 NULL == tempNode->left 时需要在链接完 node 后立即返回 root ，避免死循环。

Java Iterative

public class Solution {
  /**
   * @param root: The root of the binary search tree.
   * @param node: insert this node into the binary search tree
   * @return: The root of the new binary search tree.
   */
  public TreeNode insertNode(TreeNode root, TreeNode node) {
    // write your code here
    if (root == null) return node;
    if (node == null) return root;

    TreeNode rootcopy = root;
    while (root != null) {
      if (root.val <= node.val && root.right == null) {
        root.right = node;
        break;
      }
      else if (root.val > node.val && root.left == null) {
        root.left = node;
        break;
      }
      else if(root.val <= node.val) root = root.right;
      else root = root.left;
    }
    return rootcopy;
  }
}