LeetCode 450. 删除二叉搜索树中的节点

1、题目

2、分析

思路一：使用层序遍历。可以记录下父节点的地址。方便节点删除后，拼接父节点和子树。不过要处理的特殊情况比较多。

思路二：使用递归的处理。比较简洁、好理解。只需要处理三种情况就行（要删除的节点是叶子节点、要删除的节点只有一个子树、要删除的节点有两个子树）

3、代码

思路一：层序遍历

/**
 * 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 TreeNode deleteNode(TreeNode root, int key) {
        if(root == null) return null;
        TreeNode fatherNode = null;
        TreeNode thisNode = null;
        TreeNode leftNode = null;
        TreeNode rightNode = null;
        int leftFlag = 0;
        int rightFlag = 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        //找到要删除的节点
        while(!queue.isEmpty()){
            int size = queue.size();
            for(int i = 0; i < size; i++){
                TreeNode node = queue.poll();
                if(node.val == key){
                    thisNode = node;
                    leftNode = node.left;
                    rightNode = node.right;
                    break;
                }
                if(node.left != null){
                    queue.add(node.left);
                    if(node.left.val == key) {
                        leftFlag = 1;
                        fatherNode = node;
                    }
                }
                if(node.right != null){
                    queue.add(node.right);
                    if(node.right.val == key) {
                        rightFlag = 1;
                        fatherNode = node;
                    }
                }
            }
            if(thisNode != null) break;
        }
        //判断要删除的节点，是在父节点的左还是右
        if(thisNode == null) return root;
        if(leftFlag != 0){
            if(rightNode != null){
                fatherNode.left = rightNode;
            }
            else
            {
                fatherNode.left = leftNode;
            }
        }
        if(rightFlag != 0){
            if(rightNode != null){
                fatherNode.right = rightNode;
            }
            else
            {
                fatherNode.right = leftNode;
            }
        }
        //把要删除的节点的左子树，接到要删除节点右子树下的最左节点（也就是右子树中的最小节点）
        while(rightNode != null && rightNode.left != null){
            rightNode = rightNode.left;
        }
        if(rightNode != null) rightNode.left = leftNode;
        //特殊情况：要删除的节点是跟节点，而且只有左或者右子树
        if(fatherNode == null && thisNode.right != null) {
            return thisNode.right;
        }
        else if(fatherNode == null && thisNode.left != null){
            return thisNode.left;
        }
        //特殊情况：只有一个根节点，并且要删除的是根节点
        else if(fatherNode == null && thisNode.left == null && thisNode.right == null){return null;}
        else{
            return root;
        }
    }
}

递归解法（简洁、容易理解）：

/**
 * 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 TreeNode deleteNode(TreeNode root, int key) {
        if(root == null) return null;
        if(root.val == key){
            //该节点是叶子节点
            if(root.left == null && root.right == null) return null;
            //该节点只有一个孩子
            if(root.left == null && root.right != null) return root.right;
            if(root.left != null && root.right == null) return root.left;
            //该节点有两个孩子
            TreeNode minNode = getMinNode(root.right);
            root.val = minNode.val;
            root.right = deleteNode(root.right, minNode.val);
        }

        if(root.val > key){
            root.left = deleteNode(root.left, key);
        }
        else{
            root.right = deleteNode(root.right, key);
        }
        return root;
    }

    private TreeNode getMinNode(TreeNode root){
        while(root.left != null)
        {
            root = root.left;
        }
        return root;
    }
}