[LeetCode] 112. Path Sum ☆(二叉树是否有一条路径的sum等于给定的数)

Path Sum leetcode java

描述

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

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

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

解析

递归解法

正常的树的递归操作。

非递归，使用队列

记录每条路径的值。

代码

递归解法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (null == root) {
            return false;
        }
        if (root.val == sum && root.left == null && root.right == null) {
            return true;
        }
        boolean leftFlag = hasPathSum(root.left, sum - root.val);
        boolean rightFlag = hasPathSum(root.right, sum - root.val);
        return leftFlag || rightFlag;
    }
}

非递归，使用队列

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
 
        LinkedList<TreeNode> nodes = new LinkedList<TreeNode>();
        LinkedList<Integer> values = new LinkedList<Integer>();
 
        nodes.add(root);
        values.add(root.val);
 
        while(!nodes.isEmpty()){
            TreeNode curr = nodes.poll();
            int sumValue = values.poll();
 
            if(curr.left == null && curr.right == null && sumValue==sum){
                return true;
            }
 
            if(curr.left != null){
                nodes.add(curr.left);
                values.add(sumValue+curr.left.val);
            }
 
            if(curr.right != null){
                nodes.add(curr.right);
                values.add(sumValue+curr.right.val);
            }
        }
 
        return false;
    }
}