450. Delete Node in a BST

问题描述：

Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST.

Basically, the deletion can be divided into two stages:

1. Search for a node to remove.
2. If the node is found, delete the node.

 

Note: Time complexity should be O(height of tree).

Example:

root = [5,3,6,2,4,null,7]
key = 3

    5
   / \
  3   6
 / \   \
2   4   7

Given key to delete is 3. So we find the node with value 3 and delete it.

One valid answer is [5,4,6,2,null,null,7], shown in the following BST.

    5
   / \
  4   6
 /     \
2       7

Another valid answer is [5,2,6,null,4,null,7].

    5
   / \
  2   6
   \   \
    4   7

 

解题思路：

首先考虑是否会出现树中不含有目标节点的情况：

　　若不含目标节点，则可以直接返回root，我们可以用一个bool值来标志是否找到目标节点。

找到目标节点后，目标节点可能出现的位置：

　　1.根节点

　　2.树中

　　3.叶子节点

其实无论如何我们要做的首先是融合目标节点的子树，那么此时又有4中情况：

　　1.两个子树都不存在（即叶子节点）：返回null

　　2.只存在一个子树：存在左子树或右子树，这个时候融合结果为存在的那一课子树，则我们可以直接返回存在的子树

　　3.两个子树都存在：我的选择是将右子树加到左子树的最右的叶子节点。

此时我们有融合好的新的子树的根节点：newRoot

若我们删除的是根节点：则直接返回newRoot

其他的则可以将融合后的新子树附加到原来节点的位置上

 

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if(!root) return root;
        stack<TreeNode*> stk;
        TreeNode* cur = root;
        TreeNode* pre = NULL;
        bool findIt = false;
        while(cur || !stk.empty()){
            if(cur){
                if(cur->val == key){
                    findIt = true;
                    break;
                }
                pre = cur;
                stk.push(cur);
                cur = cur->left;
            }else{
                cur = stk.top();
                pre = cur;
                stk.pop();
                cur = cur->right;
            }
        }
        if(findIt){
            TreeNode* newRoot = UnionSubTree(cur);
            if(pre){
                if(cur->val > pre->val){
                    pre->right = newRoot;
                }else{
                    pre->left = newRoot;
                }
            }else{
                root = newRoot;
            }
        }
        return root;
    }
    TreeNode* UnionSubTree(TreeNode* root){
        if(!root->left && !root->right) return NULL;
        else if(root->left && !root->right) return root->left;
        else if(!root->left && root->right) return root->right;
        TreeNode* unionPoint = root->left;
        while(unionPoint->right) unionPoint = unionPoint->right;
        unionPoint->right = root->right;
        return root->left;
    }
};