112. Path Sum

题目：

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.

 

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 Tree Depth-first Search 

链接：http://leetcode.com/problems/path-sum/ 

题解：

求二插树中是否存在路径使得节点和为目标值。当节点左右子节点都为空时才是leaf，所以使用DFS就可以了。有空要研究一下递归的空间复杂度以及Morris Travel.

Time Complexity- O(n)， Space Complexity - O(n)。

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null)
            return false;        
        sum -= root.val;
        if(sum == 0 && root.left == null && root.right == null)
            return true;
        return hasPathSum(root.left, sum) || hasPathSum(root.right, sum);
    }
}

Update:

/**
 * Definition for a binary tree node.
 * 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;
        if(root.left == null && root.right == null && root.val == sum)
            return true;
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}

 

二刷:

方法跟一刷一样。

Java:

Time Complexity- O(n)， Space Complexity - O(n)。

/**
 * Definition for a binary tree node.
 * 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;
        }
        sum -= root.val;
        if (sum == 0 && root.left == null && root.right == null) {
            return true;
        }
        return hasPathSum(root.left, sum) || hasPathSum(root.right, sum);
    }
}

 

三刷:

Java:

/**
 * Definition for a binary tree node.
 * 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;
        }
        sum -= root.val;
        if (sum == 0 && root.left == null && root.right == null) {
            return true;
        } else {
            return hasPathSum(root.left, sum) || hasPathSum(root.right, sum);
        }
    }
}

 

测试：