【leetcode】No.113：path-sum-Ⅱ

题目描述

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 andsum = 22,

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

return

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

思路

在遍历二叉树的时候就将节点的值放入列表中，并计算权值之和，到叶子节点与sum比较，如果相等，就把列表放入结果列表中。递归非递归都可以，注意非递归需要用后序非递归改写，只有右子树为空或者已经遍历过才能把右子节点从列表中拿出来。

代码

递归实现

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.ArrayList;

public class Solution {
    ArrayList<ArrayList<Integer>> res;
    ArrayList<Integer> tmp;

    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
        res = new ArrayList();
        tmp = new ArrayList();
        preOrder(root, sum);
        return res;
    }

    private void preOrder(TreeNode root, int sum){
        if (root != null){
            sum -= root.val;
            tmp.add(root.val);
            //和为sum时把列表放入结果列表中
            if (root.left == null && root.right == null && sum == 0){
                res.add(new ArrayList<>(tmp));
            }
            else {
                preOrder(root.left, sum);
                preOrder(root.right, sum);
            }
            //遍历结束从列表中把该节点值拿出来
            tmp.remove(tmp.size() - 1);
        }
    }
}

非递归实现

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.ArrayList;
import java.util.Stack;

public class Solution {
    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<>();
        ArrayList<Integer> tmp = new ArrayList<>();
        Stack<TreeNode> stack = new Stack();
        TreeNode last = null;  //用于存储右节点是否被遍历过
        TreeNode cur = root;
        int count = 0;
        while (!stack.isEmpty() || cur != null){
            if (cur != null){
                count += cur.val;
                tmp.add(cur.val);
                //和为sum时把列表放入结果列表中
                if (count == sum && cur.left == null && cur.right == null){
                    res.add(new ArrayList(tmp));
                }
                stack.push(cur);
                cur = cur.left;
            }
            else {
                cur = stack.peek();
                //如果没有遍历过则往右子树遍历
                if (cur.right != null && last != cur.right){
                    cur = cur.right;
                }
                else {
                    cur = stack.pop();
                    //遍历过则从列表中拿出节点值，并把计算的总值减去
                    tmp.remove(tmp.size() - 1);
                    count -= cur.val;
                    //标记上一个节点
                    last = cur;
                    cur = null;
                }
            }
        }
        return res;
    }
}