538二叉搜索树转换为累加树

递归：使用pronode标记仅小于该节点的节点，使用右中左的顺序 根据pronode值修改节点值

class Solution {TreeNode pronode = null;public TreeNode convertBST(TreeNode root) {if (root==null){return root;}if (root.right!=null){convertBST(root.right);}if (pronode!=null){root.val += pronode.val;}pronode = root;if (root.left!=null){convertBST(root.left);}return root;}
}

77组合

回溯法

class Solution {List<List<Integer>> result= new ArrayList<>();LinkedList<Integer> path = new LinkedList<>();public List<List<Integer>> combine(int n, int k) {backtracking(n,k,1);return result;}public void backtracking(int n,int k,int startIndex){if (path.size() == k){result.add(new ArrayList<>(path));return;}for (int i =startIndex;i<=n;i++){path.add(i);backtracking(n,k,i+1);path.removeLast();}}
}

收获

组合问题可以转化成树从而更容易理解

树问题和回溯密不可分