LeetCode Path Sum II

标签：

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

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

return

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

思路：和上一题差不多，多个记录而已。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private List<List<Integer>> ans = new ArrayList<List<Integer>>();
	private int data[] = new int[1000];
	
	private void dfs(TreeNode root, int sum, int cur) {
		if (root == null) return;
		if (root.left == null && root.right == null && root.val == sum) {
			List<Integer> tmp = new ArrayList<Integer>();
			for (int i = 0; i < cur; i++) tmp.add(data[i]);
			tmp.add(sum);
			ans.add(tmp);
			return;
		}
		
		sum -= root.val;
		data[cur] = root.val;
		if (root.left != null) dfs(root.left, sum, cur+1);
		if (root.right != null) dfs(root.right, sum, cur+1);
	}
	
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
    	ans.clear();
    	dfs(root, sum, 0);
    	return ans;
    }
}

LeetCode Path Sum II

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原文地址：http://blog.csdn.net/u011345136/article/details/45618307