437. 路径总和 III

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原题

437. 路径总和 III

给定一个二叉树的根节点 root ，和一个整数 targetSum ，求该二叉树里节点值之和等于 targetSum 的路径的数目。

路径不需要从根节点开始，也不需要在叶子节点结束，但是路径方向必须是向下的（只能从父节点到子节点）。

输入：root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
输出：3
解释：和等于 8 的路径有 3 条，如图所示。

示例 2：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：3

root =

[1000000000,1000000000,null,294967296,null,1000000000,null,1000000000,null,1000000000]

targetSum =0

输出 0

代码思路：

由于整数溢出问题，这些大数相加会溢出：

1000000000 + 1000000000 = 2000000000 (正常)
2000000000 + 294967296 = 2294967296 > 2^31-1 = 2147483647
发生整数溢出，后续计算都不准确

/**
 * 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 pathSum(TreeNode root, int targetSum) {
        if(root==null){
            return 0;
        }
        Map<Long,Integer> map = new HashMap<>();
        map.put(0L,1);
        Long sum =0L;
        return deepSum(root,targetSum,sum,map);
    }
    public static int deepSum(TreeNode root, int targetSum, Long sum ,Map<Long,Integer> map){
       if(root!=null){
            sum+=root.val;
        }
        int count = 0;
        if(map.containsKey(sum-targetSum)){
            count+=map.get(sum-targetSum);
        }
        map.put(sum,map.getOrDefault(sum,0)+1);
        Map<Long,Integer> map1 = new HashMap<>();
        
       for(Long x:map.keySet()){
            map1.put(x,map.get(x));
           
        }
        if(root.left!=null){
            count+=deepSum(root.left,targetSum,sum,map1);
        }
        if(root.right!=(null)){
            count+=deepSum(root.right,targetSum,sum,map);
        }
        return count;
    }
}

优化更新：

/**
 * 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 pathSum(TreeNode root, int targetSum) {
        if(root==null){
            return 0;
        }
        Map<Long,Integer> map = new HashMap<>();
        map.put(0L,1);
        Long sum =0L;
        return deepSum(root,targetSum,sum,map);
    }
    public static int deepSum(TreeNode root, int targetSum, Long sum ,Map<Long,Integer> map){
       if(root!=null){
            sum+=root.val;
        }
        int count = 0;
        if(map.containsKey(sum-targetSum)){
            count+=map.get(sum-targetSum);
        }
        map.put(sum,map.getOrDefault(sum,0)+1);
        if(root.left!=null){
            count+=deepSum(root.left,targetSum,sum,map);
        }
        if(root.right!=(null)){
            count+=deepSum(root.right,targetSum,sum,map);
        }
        //退出当前节点时
        map.put(sum,map.getOrDefault(sum,0)-1);
        return count;
    }
}