LC.P450[删除二叉搜索树中的节点]

方法一：从当前节点的左子树中选择值最大的节点代替root

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/**
 * 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.right = deleteNode(root.right, key);
        else if (root.val > key) root.left = deleteNode(root.left, key);
        else {
            if (root.left == null) return root.right;
            if (root.right == null) return root.left;
            // 寻找当前节点的左子树中的最大值
            TreeNode t = root.left;
            while (t.right != null) t = t.right;
            // 将当前节点的右子树直接作为 t 的右子树
            t.right = root.right;
            return root.left;
        }
        return root;
    }
}
  • 时间复杂度：$O(h)$，$h$为树的深度
  • 空间复杂度：$O(h)$

方法二：从当前节点的右子树中选择值最小的节点代替root

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/**
 * 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.right = deleteNode(root.right, key);
        else if (root.val > key) root.left = deleteNode(root.left, key);
        else {
            if (root.left == null) return root.right;
            if (root.right == null) return root.left;
            // 寻找当前节点的右子树中的最小值
            TreeNode t = root.right;
            while (t.left != null) t = t.left;
            // 将当前节点的左子树直接作为 t 的左子树
            t.left = root.left;
            return root.right;
        }
        return root;
    }
}
  • 时间复杂度：$O(h)$，$h$为树的深度
  • 空间复杂度：$O(h)$