112. Path Sum (E)-优快云博客

本文探讨了PathSum问题，即在二叉树中寻找一条从根节点到叶子节点的路径，其上所有数值之和等于给定的目标值。通过递归回溯的方法，详细解释了如何检查这样的路径是否存在。

Path Sum (E)

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.

Note: A leaf is a node with no children.

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.

题意

判断给定树中是否存在这样一条从根到叶的路径，使得路径上的数字之和等于目标值。

思路

回溯法。

代码实现

/**
 * 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) {
        return hasPathSum(root, 0, sum);
    }

    private boolean hasPathSum(TreeNode x, int preSum, int sum) {
        if (x == null) {
            return false;
        }
        // 路径数字和等于目标值且当前结点为叶结点，说明找到了
        if (preSum + x.val == sum && x.left == null & x.right == null) {
            return true;
        }


        if (hasPathSum(x.left, preSum + x.val, sum)) {
            return true;
        }
        if (hasPathSum(x.right, preSum + x.val, sum)) {
            return true;
        }

        return false;
    }
}