如何快速备案网站专业培训

617. 合并二叉树

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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 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

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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 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

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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 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

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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 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

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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 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

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/*** 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;}}}