代码随想录三刷day24

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前言

涉及到二叉树的构造，无论普通二叉树还是二叉搜索树一定前序，都是先构造中节点。求普通二叉树的属性，一般是后序，一般要通过递归函数的返回值做计算。求二叉搜索树的属性，一定是中序了，要不白瞎了有序性了。

一、力扣450. 删除二叉搜索树中的节点

在这里插入代码片/**
 * 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){
            root.left = deleteNode(root.left,key);
        }else if(root.val < key){
            root.right = deleteNode(root.right,key);
        }else{
            if(root.left == null && root.right == null){
                return null;
            }else if(root.left == null && root.right != null){
                return root.right;
            }else if(root.left != null && root.right == null){
                return root.left;
            }else{
                TreeNode p = root.right;
                while(p.left != null){
                    p = p.left;
                }
                p.left = root.left;
                root = root.right;
                return root;
            }
        }
        return root;
    }
}

二、力扣669. 修剪二叉搜索树

在这里插入代码片/**
 * 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 trimBST(TreeNode root, int low, int high) {
        if(root == null){
            return null;
        }
        if(root.val < low){
            return trimBST(root.right,low,high);
        }else if(root.val > high){
            return trimBST(root.left,low,high);
        }else{
            root.left = trimBST(root.left,low,high);
            root.right = trimBST(root.right,low,high);
        }
        return root;
    }
}

三、力扣108. 将有序数组转换为二叉搜索树

在这里插入代码片/**
 * 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 sortedArrayToBST(int[] nums) {
        return fun(nums,0,nums.length-1);
    }
    public TreeNode fun(int[] nums, int low, int high){
        if(low > high){
            return null;
        }
        int mid = low + (high-low)/2;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = fun(nums,low,mid-1);
        root.right = fun(nums,mid+1,high);
        return root;
    }
}

四、力扣538. 把二叉搜索树转换为累加树

/**
 * 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 {
    int sum = 0;
    public TreeNode convertBST(TreeNode root) {
        if(root == null){
            return null;
        }
        root.right = convertBST(root.right);
        root.val = sum + root.val;
        sum = root.val;
        root.left = convertBST(root.left);
        return root;
    }
}