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113. 路径总和 II.js
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113. 路径总和 II.js
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/**
113. 路径总和 II
给定一个二叉树和一个目标和,找到所有从根节点到叶子节点路径总和等于给定目标和的路径。
说明: 叶子节点是指没有子节点的节点。
示例:
给定如下二叉树,以及目标和 sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
返回:
[
[5,4,11,2],
[5,8,4,5]
]
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @param {number} sum
* @return {number[][]}
*/
var pathSum = function (root, sum) {
if (!root) return [];
let res = [];
dfs(root, sum, [], res);
return res;
};
function dfs(root, sum, solution, res) {
if (!root) return;
solution.push(root.val);
if (!root.left && !root.right && root.val === sum) {
res.push([...solution]);
}
dfs(root.left, sum - root.val, solution, res);
dfs(root.right, sum - root.val, solution, res);
solution.pop();
}