剑指offer-面试题25：二叉树中和为某一值的路径

题目

class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
public TreeNode() {
// TODO Auto-generated constructor stub
}

解

package Chapter4;

import java.util.Stack;

public class FindPathInBinaryTree {

public void findPath(TreeNode root,int expectedSum){
if(root==null)
return;

Stack<Integer> path=new Stack<Integer>();
int currentSum=0;
findPath2(root,expectedSum,path,currentSum);

}
public void findPath2(TreeNode root, int expectedSum, Stack<Integer> path, int currentSum) {
// TODO Auto-generated method stub
currentSum+=root.val;
path.push(root.val);

//如果是叶结点，并且路径上结点的和等于输入的值
//打印出这条路径
boolean isLeaf=root.left==null&&root.right==null;
if(currentSum==expectedSum&&isLeaf){
System.out.print("A path is found:");
for(Integer i:path){
System.out.print(i+"  ");
}
System.out.println("");
}

//如果不是叶结点，则遍历它的子结点
if(root.left!=null){
findPath2(root.left,expectedSum,path,currentSum);
}
if(root.right!=null){
findPath2(root.right,expectedSum,path,currentSum);
}

//在返回到父结点之前，在路径上删除当前结点
currentSum-=root.val;
path.pop();

}
public static void main(String[] args) {
// TODO Auto-generated method stub
TreeNode root=new TreeNode(10);
TreeNode Node5=new TreeNode(5);
TreeNode Node12=new TreeNode(12);
TreeNode Node4=new TreeNode(4);
TreeNode Node7=new TreeNode(7);

root.left=Node5;
root.right=Node12;
Node5.left=Node4;
Node5.right=Node7;

FindPathInBinaryTree fP=new FindPathInBinaryTree();
fP.findPath(root, 22);
}
}

class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
public TreeNode() {
// TODO Auto-generated constructor stub
}
}