代码随想录day22

一、二叉搜索树的最近公共祖先（LeetCode235）

递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null)    return null;
        if(root.val > p.val && root.val > q.val){
            TreeNode left = lowestCommonAncestor(root.left,p,q);
            if(left != null)    return left;
        }else if(root.val < p.val && root.val < q.val){
            TreeNode right = lowestCommonAncestor(root.right,p,q);
            if(right != null)   return right;
        }
        return root;
    }
}

迭代

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        while(root != null){
            if(root.val > p.val && root.val > q.val){
                root = root.left;
            }else if(root.val < p.val && root.val < q.val){
                root = root.right;
            }else{
                return root;
            }
        }
        return null;
    }
}

二、二叉搜索树中的插入操作（LeetCode701）

递归

/**
 * 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 insertIntoBST(TreeNode root, int val) {
        if(root == null){
            TreeNode node = new TreeNode(val);
            return node;
        }
        if(val < root.val){
            root.left = insertIntoBST(root.left,val);
        }
        if(val > root.val){
            root.right = insertIntoBST(root.right,val);
        }
        return root;
    }
}

迭代

/**
 * 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 insertIntoBST(TreeNode root, int val) {
        if(root == null){
            TreeNode node = new TreeNode(val);
            return node;
        }
        TreeNode newRoot = root;
        TreeNode pre = root;
        while (root != null) {
            pre = root;
            if (root.val > val) {
                root = root.left;
            } else if (root.val < val) {
                root = root.right;
            } 
        }
        if (pre.val > val) {
            pre.left = new TreeNode(val);
        } else {
            pre.right = new TreeNode(val);
        }
        return newRoot;
    }
}

三、删除二叉搜索树中的节点（LeetCode450）

递归

/**
 * 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){
            if(root.left == null && root.right == null){
                return null;
            }else if(root.left != null && root.right == null){
                return root.left;    
            }else if(root.left == null && root.right != null){
                return root.right;
            }else{
                TreeNode node = root.right;
                while(node.left != null){
                    node = node.left;
                }
                node.left = root.left;
                return root.right;
            }
        }
        if(root.val > key){
            root.left = deleteNode(root.left,key);
        }
        if(root.val < key){
            root.right = deleteNode(root.right,key);
        }
        return root;
    }
}