本文介绍了一种解决LeetCode中路径总和II问题的方法。该问题要求找到二叉树中所有从根节点到叶子节点的路径，并且这些路径上的节点值之和等于给定的目标值。通过深度优先搜索(DFS)算法实现解决方案。
Binary Tree的一个基本题。
遍历到底，比较sum vs. target。
注意divide的情况。要把遍历的例子写写。
LeetCode: Path Sum II
```
/*
Given a binary tree, find all paths that sum of the nodes in the path equals to a given number target.
A valid path is from root node to any of the leaf nodes.
Example
Given a binary tree, and target = 5:
1
/ \
2 4
/ \
2 3
return
[
[1, 2, 2],
[1, 4]
]
Tags Expand
Binary Tree Binary Tree Traversal
*/
/*
Thoughts:
path: has to be from root to leaf.
binary tree: no order logic in the tree.
DPS on all nodes. If final sum == target, add list of nodes into rst
*/
public class Solution {
public class TreeNode {
public int val;
public TreeNode left, right;
public TreeNode(int val) {
this.val = val;
this.left = this.right = null;
}
}
/**
* @param root the root of binary tree
* @param target an integer
* @return all valid paths
*/
public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {
List<List<Integer>> rst = new ArrayList<List<Integer>>();
if (root == null) {
return rst;
}
dfs(rst, new ArrayList<Integer>(), root, 0, target);
return rst;
}
private void dfs(List<List<Integer>> rst, ArrayList<Integer> list, TreeNode root, int add, int sum) {
list.add(root.val);
if (root.left == null && root.right == null) {
if (add + root.val == sum) {
rst.add(new ArrayList<Integer>(list));
}
return;
}
if (root.left != null) {
dfs(rst, list, root.left, add + root.val, sum);
list.remove(list.size() - 1);
}
if (root.right != null) {
dfs(rst, list, root.right, add + root.val, sum);
list.remove(list.size() - 1);
}
}
}
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