题目
结果
代码
逃课做法
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| 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) {
List<Integer> path = new LinkedList<>();
dfs(root, path);
path.add(val);
path.sort(Integer::compareTo);
return buildTree(path);
}
private TreeNode buildTree(List<Integer> path) {
TreeNode root = new TreeNode(0);
TreeNode p = root;
for (int val : path) {
p.right = new TreeNode(val);
p = p.right;
}
return root.right;
}
private void dfs(TreeNode root, List<Integer> path) {
if (root == null) {
return;
}
path.add(root.val);
dfs(root.left, path);
dfs(root.right, path);
}
}
|
这题目不管是用迭代还是递归都挺简单的，好像也没必要逃课。
一般做法
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| class Solution {
public TreeNode insertIntoBST(TreeNode root, int val) {
if (root == null) {
return new TreeNode(val);
}
TreeNode p = root;
while (true) {
if (val < p.val) {
if (p.left == null) {
p.left = new TreeNode(val);
break;
} else {
p = p.left;
}
} else {
if (p.right == null) {
p.right = new TreeNode(val);
break;
} else {
p = p.right;
}
}
}
return root;
}
}
|
复杂度
时间复杂度：O（n），最坏情况下
空间复杂度：O（1），迭代
