【leetcode】112. Path Sum【java】

最新推荐文章于 2021-12-31 16:11:03 发布

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本文介绍了一种算法，用于判断二叉树中是否存在从根节点到叶子节点的路径，使得这条路径上所有节点值之和等于给定的数值。通过递归和非递归两种方法实现了这一功能。

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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.

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

//非递归：DFS  使用栈
public class  Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        Stack<TreeNode> stack = new Stack<TreeNode>();
        Stack<Integer> sums = new Stack<Integer>();
        stack.push(root);
        sums.push(sum);
        while(!stack.isEmpty()) {
            TreeNode node = stack.pop();
            int value = sums.pop();
            if (node.left == null && node.right == null && node.val == value) {
                return true;
            }
            if (node.left != null) {
                stack.push(node.left);
                sums.push(value - node.val);
            }
            if (node.right != null){
                stack.push(node.right);
                sums.push(value - node.val);
            }
        }
        return false;
    }
}