*LeetCode- Sum Root to Leaf Numbers

这个题一开始一直想从叶子开始加，后来发现从root向下走比较方便。递归函数传一个记录到达当前节点的和。

recursive：

public class Solution {
public int sumNumbers(TreeNode root) {
return sum(root,0);
}
public int sum (TreeNode root, int sumNow){
if ( root == null )
return 0;
if ( root.left == null && root.right == null )
return sumNow = sumNow*10 + root.val;
return sum(root.left, sumNow*10 + root.val) + sum (root.right, sumNow * 10 + root.val);

}
}注意判断的是root是否为空，是否为叶子

iterative：stack实现的dfs 这个应该就是bottom up的

public static int sumNumbers(TreeNode root) {
int sum = 0, num = 0;
TreeNode lastVisited = null;
Stack<TreeNode> s = new Stack<>();
while (root != null || !s.isEmpty()) {
if (root != null) {
s.push(root);
num = num * 10 + root.val;
root = root.left;
} else {
TreeNode peek = s.peek();
if (peek.right != lastVisited && peek.right != null) {
root = peek.right;
} else {
if (peek.left == null && peek.right == null) {
sum += num;
}
s.pop();
num /= 10;
lastVisited = peek;
}
}
}
return sum;

}