二叉树day12

701. 二叉搜索树中的插入操作

//递归遍历搜索树
class Solution {
    public TreeNode insertIntoBST(TreeNode root, int val) {
        if (root == null) return new TreeNode(val);
        if (val > root.val) root.right = insertIntoBST(root.right, val);
        else if(val < root.val) root.left = insertIntoBST(root.left, val);
        return root;
    }
}

//迭代
class Solution {
    public TreeNode insertIntoBST(TreeNode root, int val) {
        if (root == null) return new TreeNode(val);
        TreeNode cur = root, pre = root;
        while (cur != null) {
            pre = cur;
            if (cur.val > val) cur = cur.left;
            else cur = cur.right;
        }
        if (pre.val > val) pre.left = new TreeNode(val);
        else pre.right = new TreeNode(val);
        return root;
    }
}

450. 删除二叉搜索树中的节点

/* 递归
 * 如果找到key则重构，所有可能出现的情况如下
 * 1.有一个子树为空，直接返回不空的子树
 * 2.子树都空返回null
 * 3.子树都不为空，删除根节点， 左子树作为右子树最左子结点的左子树
 */
class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if (root == null) return null;
        //根节点为key则重构
        else if (root.val == key) {
            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 cur = root.right;
                while (cur.left != null) {
                    cur = cur.left;
                }
                cur.left = root.left;
                return root.right;
            }
        }
        //没有找到 根据搜索树的特点找key
        if (root.val > key) root.left = deleteNode(root.left, key);
        else if(root.val < key) root.right = deleteNode(root.right, key);
        return root;
    }
}

迭代法

//迭代
class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if (root == null) return root;
        TreeNode parent = null,  cur = root;
        while (cur != null) {
            if (cur.val == key) break;
            parent = cur;
            if (cur.val > key) cur = cur.left;
            else cur = cur.right;
        }
        //根val为key
        if (parent == null) root = rebuild(cur);
        else if (parent.left != null && parent.left.val == key) parent.left = rebuild(cur);
        else if (parent.right != null && parent.right.val == key) parent.right = rebuild(cur);

        return root;
    }
    private TreeNode rebuild(TreeNode root) {
        //没找到
        if (root == null) return null;
        else if (root.right == null) return root.left;
        //将左子树作为右子树最左子结点的左子树 包含了左子树为空的情况
        TreeNode cur  = root.right;
        while (cur.left != null) {
            cur = cur.left;
        }
        cur.left = root.left;
        return root.right;
    }
}

669. 修剪二叉搜索树

//递归
class Solution {
    public TreeNode trimBST(TreeNode root, int low, int high) {
        if (root == null) return null;
        if (root.val < low) {
            if (root.right == null) return null;
            //继续修剪右子树
            else return trimBST(root.right, low, high);
        } else if (root.val > high) {
            if (root.left == null) return null;
            //继续修剪左子树
            else return trimBST(root.left, low, high);
        }

        //小于low是要被裁剪的
        if (root.val >= low) root.left = trimBST(root.left, low, high);
        if (root.val <= high) root.right = trimBST(root.right, low, high);
        return root;
    }
}

//迭代
class Solution {
    public TreeNode trimBST(TreeNode root, int low, int high) {
        if (root == null) return null;
        //找到位于[low, high]区间内的结点作为根节点，然后修剪子树
        while (root != null && (root.val > high || root.val < low)) {
            if (root.val > high) root = root.left;
            else root = root.right;
        }
        //修剪子树
        TreeNode cur = root;
        //修剪左子树中小于low的部分
        while (cur != null) {
            while (cur.left != null && cur.left.val < low) {
                cur.left = cur.left.right;
            }
            cur = cur.left;
        }
        cur = root;
        //修剪右子树大于high的部分
        while (cur != null) {
            while (cur.right != null && cur.right.val > high) {
                cur.right = cur.right.left;
            }
            cur = cur.right;
        }
        return root;
    }
}

参考：programmercarl.com