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;
}
}