import java.util.Stack;
/**
*
* Source : https://oj.leetcode.com/problems/sum-root-to-leaf-numbers/
*
*
* Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
* An example is the root-to-leaf path 1->2->3 which represents the number 123.
*
* Find the total sum of all root-to-leaf numbers.
*
* For example,
*
* 1
* / \
* 2 3
*
* The root-to-leaf path 1->2 represents the number 12.
* The root-to-leaf path 1->3 represents the number 13.
*
* Return the sum = 12 + 13 = 25.
*
*/
public class SumRootToLeafNumbers {
/**
* 求出由根节点到叶子节点组成所有数字的和
*
* 可以使用深度优先DFS求出所有的数字然后求和
* 也可以使用BFS,逐层求和,求和的时候不是直接加该节点的值,是该节点的值加上上一节点的10倍(子节点处表示的数字就是root.value*10 + node.value)
* 直到最后一个没有子节点的节点的时候,该节点的值就是最后的和
*
* 相当于将根节点到叶子节点路径所表示的数字集中到叶子节点上,然后对叶子节点求和
*
* @param root
* @return
*/
public int sum (TreeNode root) {
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
int sum = 0;
while (stack.size() > 0) {
TreeNode node = stack.pop();
if (node.leftChild != null) {
node.leftChild.value += node.value * 10;
stack.push(node.leftChild);
}
if (node.rightChild != null) {
node.rightChild.value += node.value * 10;
stack.push(node.rightChild);
}
if (node.leftChild == null && node.rightChild == null) {
sum += node.value;
}
}
return sum;
}
public TreeNode createTree (char[] treeArr) {
TreeNode[] tree = new TreeNode[treeArr.length];
for (int i = 0; i < treeArr.length; i++) {
if (treeArr[i] == '#') {
tree[i] = null;
continue;
}
tree[i] = new TreeNode(treeArr[i]-'0');
}
int pos = 0;
for (int i = 0; i < treeArr.length && pos < treeArr.length-1; i++) {
if (tree[i] != null) {
tree[i].leftChild = tree[++pos];
if (pos < treeArr.length-1) {
tree[i].rightChild = tree[++pos];
}
}
}
return tree[0];
}
private class TreeNode {
TreeNode leftChild;
TreeNode rightChild;
int value;
public TreeNode(int value) {
this.value = value;
}
public TreeNode() {
}
}
public static void main(String[] args) {
SumRootToLeafNumbers sumRootToLeafNUmbers = new SumRootToLeafNumbers();
char[] arr = new char[]{'1','2','3'};
System.out.println(sumRootToLeafNUmbers.sum(sumRootToLeafNUmbers.createTree(arr)));
}
}