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前言

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二叉树的递归分为「遍历」和「分解问题」两种思维模式，这道题可以同时用到两种思维模式。 「遍历」的话很简单，你对 BST 做中序遍历，其结果就是有序的，重新构造出题目要求的这个类似链表的二叉树即可。

一、力扣113. 路径总和 II

/*** 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 {List<List<Integer>> res = new ArrayList<>();List<Integer> path = new ArrayList<>();public List<List<Integer>> pathSum(TreeNode root, int targetSum) {if(root == null){return new ArrayList<>();}fun(root,targetSum,0);return res;}public void fun(TreeNode root,int targetSum, int sum){if(root == null){return;}path.add(root.val);if(root.left == null && root.right == null){if(sum + root.val == targetSum){res.add(new ArrayList<>(path));}path.remove(path.size()-1);return;}fun(root.left,targetSum,sum+root.val);fun(root.right,targetSum,sum+root.val);path.remove(path.size()-1);}
}

二、力扣1430. 判断给定的序列是否是二叉树从根到叶的路径

/*** 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 {int len = 0;boolean flag = false;public boolean isValidSequence(TreeNode root, int[] arr) {len = arr.length-1;fun(root,arr,0);return flag;}public void fun(TreeNode root, int[] arr , int index){if(root == null || index > len){return;}if(root.val == arr[index]){if(index == len && root.left == null && root.right == null){flag = true;}fun(root.left,arr,index+1);fun(root.right,arr,index+1);}return;}
}

三、力扣897. 递增顺序搜索树

/*** 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 {TreeNode pre = null, first = null;boolean flag = true;public TreeNode increasingBST(TreeNode root) {fun(root);pre.left = null;pre.right = null;return first;}public void fun(TreeNode root){if(root == null){return;}fun(root.left);if(pre != null){pre.right = root;pre.left = null;}pre = root;if(flag){first = root;flag = false;}fun(root.right);}
}

四、力扣938. 二叉搜索树的范围和

/*** 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 {int sum = 0;public int rangeSumBST(TreeNode root, int low, int high) {fun(root,low,high);return sum;}public void fun(TreeNode root, int low, int high){if(root == null){return ;}if(root.val > high){fun(root.left,low,high);}else if(root.val < low){fun(root.right,low,high);}else{sum += root.val;fun(root.left,low,high);fun(root.right,low,high);}}
}