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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，这道题需要用到「遍历」的思维模式

一、力扣1602. 找到二叉树中最近的右侧节点

/*** 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 dif = Integer.MAX_VALUE;TreeNode res = null;int high = -1;int flag = -1;public TreeNode findNearestRightNode(TreeNode root, TreeNode u) {fun(root, u, 1,1);return res;}public void fun(TreeNode root, TreeNode u, int index, int depth){if(root == null){return;}if(root == u){high = depth;flag = index;}else if(high == depth){if((index - flag) < dif){res = root;dif = index - flag;}}fun(root.left , u, index * 2, depth + 1);fun(root.right, u, index *2 + 1, depth + 1);}
}

二、力扣437. 路径总和 III

/*** 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 {Map<Long,Integer> preSumPath = new HashMap<>();int res = 0;long targetSum , pathSum ;public int pathSum(TreeNode root, int targetSum) {if(root == null){return 0;}this.targetSum = targetSum;this.pathSum = 0;preSumPath.put(0L,1);fun(root);return res;}public void fun(TreeNode root){if(root == null){return;}pathSum += root.val;res += preSumPath.getOrDefault(pathSum - targetSum,0);preSumPath.put(pathSum, preSumPath.getOrDefault(pathSum,0)+1);fun(root.left);fun(root.right);preSumPath.put(pathSum,preSumPath.getOrDefault(pathSum,0)-1);pathSum -= root.val;}
}

三、力扣560. 和为 K 的子数组

class Solution {public int subarraySum(int[] nums, int k) {int[] preSum = new int[nums.length+1];for(int i = 0; i < nums.length; i ++){preSum[i+1] = nums[i] + preSum[i];}int count = 0;for(int low = 0; low < preSum.length - 1; low ++){for(int high = low; high < preSum.length-1; high ++){if(preSum[high + 1] - preSum[low] == k){count ++;}}}return count;}
}