LeetCode-701. Insert into a Binary Search Tree

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:

         4
       /   \
      2     7
     / \   /
    1   3 5

This tree is also valid:

         5
       /   \
      2     7
     / \   
    1   3
         \
          4

题目:将某个值插入二分查找树(BST)中,形成的新树可以有两种形式。

思路:为要插入的值建议一个newNode,如果原来的树是空树,直接返回newNode。将val与root->val比较,如果valval,如果左子树为空,则newNode为左子节点,否则在左子树上进一步如上的操作。同理,右子树也是一样的。代码如下:

class Solution {
public:
    TreeNode* insertIntoBST(TreeNode* root, int val) {
        TreeNode* newNode = (TreeNode*)malloc(sizeof(TreeNode));  //为val新建一个节点
        newNode->val = val;
        newNode->left = NULL;
        newNode->right = NULL;
		if(!root){  //如果是个空树
			return newNode;
		}
		TreeNode* p = root;
		while(p){
			if(p->val>val){
				if(p->left==NULL){
					p->left=newNode;
					break;  //如果插入成功之后,break直接跳出循环,否则是个死循环
				}
				else
					p=p->left;
			}
			else{
				if(p->right==NULL){
					p->right=newNode;
					break;   //如果插入成功之后,break直接跳出循环,否则是个死循环
				}
                else
					p=p->right;	
			}
		}
		return root;
    }
};

AC,beats 98.7%

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