代码随想三刷二叉树篇4
617. 合并二叉树
题目
代码
/**
* 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 mergeTrees(TreeNode root1, TreeNode root2) {
if(root1==null&&root2==null){
return null;
}
if(root1==null){
return root2;
}
if(root2==null){
return root1;
}
root1.val += root2.val;
TreeNode left = mergeTrees(root1.left,root2.left);
TreeNode right = mergeTrees(root1.right,root2.right);
root1.left = left;
root1.right = right;
return root1;
}
}
700. 二叉搜索树中的搜索
题目
代码
/**
* 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 searchBST(TreeNode root, int val) {
if(root==null){
return null;
}
if(root.val==val){
return root;
}
if(val>root.val){
return searchBST(root.right,val);
}else{
return searchBST(root.left,val);
}
}
}
98. 验证二叉搜索树
题目
代码
/**
* 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 boolean isValidBST(TreeNode root) {
traverse(root);
return isValidBST;
}
boolean isValidBST = true;
TreeNode pre = null;
public void traverse(TreeNode root){
if(root==null||!isValidBST){
return;
}
traverse(root.left);
if(pre!=null&&pre.val>=root.val){
isValidBST = false;
}
pre = root;
traverse(root.right);
}
}
530. 二叉搜索树的最小绝对差
题目
代码
/**
* 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 int getMinimumDifference(TreeNode root) {
traverse(root);
return minNum==Integer.MAX_VALUE?0:minNum;
}
TreeNode pre = null;
int minNum = Integer.MAX_VALUE;
public void traverse(TreeNode root){
if(root==null){
return;
}
traverse(root.left);
if(pre!=null){
minNum = Math.min(minNum,root.val-pre.val);
}
pre = root;
traverse(root.right);
}
}
501. 二叉搜索树中的众数
题目
代码
/**
* 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 int[] findMode(TreeNode root) {
traversal(root);
pq.offer(new int[]{cur,count});//最后一次没有添加
List<Integer> result = new ArrayList();
int maxCount = 0;
while(!pq.isEmpty()){
int[] temp = pq.poll();
if(temp[1]>=maxCount){
maxCount = temp[1];
result.add(temp[0]);
}else{
break;
}
}
return result.stream().mapToInt(i->i).toArray();
}
PriorityQueue<int[]> pq = new PriorityQueue<>((e1,e2)->e2[1]-e1[1]);
int cur =0;
int count = 0;
TreeNode pre = null;
public void traversal(TreeNode root){
if(root==null){
return;
}
traversal(root.left);
if(pre==null){
cur = root.val;
count++;
}else{
if(cur == root.val){
count++;
}else{
pq.offer(new int[]{cur,count});
cur = root.val;
count = 1;
}
}
pre = root;
traversal(root.right);
}
}
236. 二叉树的最近公共祖先
题目
代码
/**
* 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||root==p||root==q){
return root;
}
TreeNode left = lowestCommonAncestor(root.left,p,q);
TreeNode right = lowestCommonAncestor(root.right,p,q);
if(left==null&&right==null){
return null;
}else if(left!=null&&right==null){
return left;
}else if(right!=null&&left==null){
return right;
}else {
return root;
}
}
}