112 Path Sum
Question:
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.
Lesson Learned:
Path题可以把path的sum在变量里变化,这样,当path发生变化的时候,sum随之发生变化。如果使用外部变量,则外部变量不在recursion中,需要人工手动加减,非常不方便。
return值:如果是return一个path上的值,通常是||, Math.min, Math.max等等,如果是所有path相加,通常是+,-等等
||: 两根遍历有符合值即可 Math.min, Math.max: 两根遍历找最大/最小
recursive的变化值可以是传递函数,亦可以是return value。判断是否为真的时候用传递函数变换,用int的时候使用返回值变换。
使用外部变量一般是整个recursion是单向的变化,比如说只是增大或是只是减小,否则就要在recursion里变换很多次。
(112)本题的boolean和symmetric的boolean并不一样,symmetric的boolean是每一步都在找boolean,因此用返回参数很好。这道题是寻找有没有这样一个值,因此用外部参数比较好。
因为是path问题,所以用了一个pathSum传递函数来跟踪path的值。
用了path问题的template
Analysis:
First Attempt:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private int total;
private boolean path;
public boolean hasPathSum(TreeNode root, int sum) {
total = 0;
path = false;
helper(root, sum);
return path;
}
private void helper (TreeNode root, int sum) {
if (root == null) {
return;
}
total += root.val;
if (root.left == null && root.right == null) {
if (sum == total) {
path = true;
}
total -= root.val;
return;
}
helper(root.left, sum);
helper(root.right, sum);
total -= root.val;
}
}
https://leetcode.com/discuss/10456/accepted-my-recursive-solution-in-java
public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if(root == null) return false;
if(root.left == null && root.right == null && sum - root.val == 0) return true;
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
}
updated (12.19.15)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
// (112)本题的boolean和symmetric的boolean并不一样,symmetric的boolean是每一步都在找boolean,因此用返回参数很好。这道题是寻找有没有这样一个值,因此用外部参数比较好。
// 因为是path问题,所以用了一个pathSum传递函数来跟踪path的值。
// 用了path问题的template
private boolean hasSum;
public boolean hasPathSum(TreeNode root, int sum) {
hasSum = false;
if (root == null) {
return false;
}
int pathSum = 0;
helper(root, sum, pathSum);
return hasSum;
}
private void helper(TreeNode root, int sum, int pathSum) {
if (root == null) {
return;
}
if (root.left == null && root.right == null) {
if (pathSum + root.val == sum) {
hasSum = true;
}
}
if (root.left == null) {
helper(root.right, sum, pathSum + root.val);
}
if (root.right == null) {
helper(root.left, sum, pathSum + root.val);
}
helper(root.left, sum, pathSum + root.val);
helper(root.right, sum, pathSum + root.val);
}
}