• 【LeetCode】Path Sum II 二叉树递归


    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]
    ]
    

     

     

    Discuss

    java code : 
    /**
     * Definition for binary tree
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
            // IMPORTANT: Please reset any member data you declared, as
            // the same Solution instance will be reused for each test case.
            ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
    		if(root == null)
    			return res;
    		ArrayList<Integer> tmp = new ArrayList<Integer>();
    		recursion(root, res, tmp, sum);
    		tmp = null;
    		return res;
        }
        public void recursion(TreeNode root, ArrayList<ArrayList<Integer>> res, ArrayList<Integer> tmp, int sum)
    	{
    		if(root.left == null && root.right == null)
    		{
    			tmp.add(root.val);
    			int sum1 = 0;
    			for(int i = 0; i < tmp.size(); i++)
    			{
    				sum1 += tmp.get(i);
    			}
    			if(sum1 == sum)
    				res.add(new ArrayList<Integer>(tmp));
    			return ;
    		}
    		tmp.add(root.val);
    		if(root.left != null)
    		{	
    			recursion(root.left, res, tmp, sum);
    			tmp.remove(tmp.size() - 1);
    		}
    		if(root.right != null)
    		{
    			recursion(root.right, res, tmp, sum);
    			tmp.remove(tmp.size() - 1);
    		}
    	}
    }


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  • 原文地址:https://www.cnblogs.com/keanuyaoo/p/3400358.html
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