Path Sum II

原创于 2012-12-09 12:52:50 发布 · 1.5k 阅读

0 ·

CC 4.0 BY-SA版权

二叉树 (binary tree) 专栏收录该内容

34 篇文章

订阅专栏

本文介绍了一种算法，用于在给定的二叉树中找到所有从根节点到叶子节点的路径，使得这些路径的元素之和等于指定的数值。通过实例演示了如何实现这一算法。

Question:

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and

sum
 = 22

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

Code:

public static void sumAll(Node root, ArrayList<Integer> list, ArrayList<ArrayList<Integer>> listAll) {
		if (root == null) return;
		list.add(root.v);
		if (root.leftChild == null && root.rightChild == null) {
			listAll.add(new ArrayList<Integer>(list));
		}
		sumAll(root.leftChild, list, listAll);
		sumAll(root.rightChild, list, listAll);
		list.remove(list.size() - 1);
	}
	
	class Node {
	    Node leftChild = null;
	    Node rightChild = null;
	    int v;

	    Node(int value) {
	        this.v = value;
	    }
	}