437\. 路径总和 III

# 437. 路径总和 III (opens new window)

# Description

Difficulty: 中等

Related Topics: (opens new window), 深度优先搜索 (opens new window), 二叉树 (opens new window)

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)。

示例 1:

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。
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3

示例 2:

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

  • 二叉树的节点个数的范围是 [0,1000]
  • -109 <= Node.val <= 109
  • -1000 <= targetSum <= 1000

# Solution

Language: JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} targetSum
 * @return {number}
 */

// 方法一:深度优先搜索
var pathSum = function(root, targetSum) {
    if (root == null) {
        return 0;
    }
    
    let ret = rootSum(root, targetSum);
    ret += pathSum(root.left, targetSum);
    ret += pathSum(root.right, targetSum);
    return ret;
};

const rootSum = (root, targetSum) => {
    let ret = 0;

    if (root == null) {
        return 0;
    }
    const val = root.val;
    if (val === targetSum) {
        ret++;
    } 

    ret += rootSum(root.left, targetSum - val);
    ret += rootSum(root.right, targetSum - val);
    return ret;
}


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DFS

var pathSum = function (root, targetSum) {
    let cnt = 0;
    function dfs(node, sum, flag) {
        if (!node) return;
        sum += node.val;
        if (sum === targetSum) cnt++;
        dfs(node.left, sum, true);
        dfs(node.right, sum, true);
        if (flag) return;
        dfs(node.left, 0, false);
        dfs(node.right, 0, false);
    }
    dfs(root, 0, false);
    return cnt;
};


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var pathSum = (root, targetSum) => {
    // 深度优先遍历
    // 参数:(当前节点,当前节点前缀和)
    const dfs = (root, sum) => {
        if (!root) return 0;
        map.set(sum, (map.get(sum) || 0) + 1);
        const newSum = sum + root.val;
        // 寻找有无(newSum - targetSum)
        res += map.get(newSum - targetSum) || 0;
        // 向下遍历
        dfs(root.left, newSum);
        dfs(root.right, newSum);
        // 避免相同节点值被重复计算,确保算出的sum是从上到下的一条路径
        map.set(sum, map.get(sum) - 1);
    };

    // map中,key存的是前缀和,value存的是前缀和的数目
    const map = new Map();
    let res = 0;
    dfs(root, 0);
    return res;
};
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上次更新: 2022/8/25 下午2:49:04