有一棵二叉树，树上每个点标有权值，权值各不相同，请设计一个算法算出权值最大的叶节点到权值最小的叶节点的距离。二叉树每条边的距离为1，一个节点经过多少条边到达另一个节点为这两个节点之间的距离。
给定二叉树的根节点root，请返回所求距离。
import java.util.*;
public class Tree {
public static void main(String[] args) {
// TODO Auto-generated method stub
Tree tree = new Tree();
TreeNode root = new TreeNode(5);
TreeNode left = new TreeNode(4);
TreeNode right = new TreeNode(6);
root.left = left;
root.right = right;
TreeNode r2 = new TreeNode(8);
right.right = r2;
Map<TreeNode, TreeNode> map=new HashMap<>();
System.out.println(tree.getDis(root));
//dfs
/*tree.dfs(root, map, 1);
System.out.println(tree.dpest);
for(int i=0;i<tree.leaves.size();i++){
tree.showPath(tree.leaves.get(i),map);
}*/
//bfs
//tree.bfs(root);
}
private void showPath(TreeNode node,Map<TreeNode, TreeNode> map){
while(map.containsKey(node)){
System.out.print(node.val+"<--");
node = map.get(node);
}
System.out.println(node.val);
}
public int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE;
public int getDis(TreeNode root) {
// write code here
HashMap<Integer, Integer> map = new HashMap<>();
helper(root, map);
// showMap(map);
System.out.println("min:" + min + ",max" + max);
// root.left=null;
List<Integer> path1 = getPath(min, map, new ArrayList<Integer>());
List<Integer> path2 = getPath(max, map, new ArrayList<Integer>());
int len1 = path1.size();
int len2 = path2.size();
int result = 0;
for (int i = len1 - 1, j = len2 - 1; i >= 0 && j >= 0; i--, j--) {
if (path1.get(i) == path2.get(j)) {
result = i + j;
}
}
return result;
}
private List<Integer> getPath(int i, Map<Integer, Integer> map, List<Integer> list) {
while (map.containsKey(i)) {
list.add(i);
i = map.get(i);
}
list.add(i);
return list;
}
void showMap(Map<Integer, Integer> map) {
Iterator<Integer> iterator = map.keySet().iterator();
while (iterator.hasNext()) {
Integer i = iterator.next();
System.out.println("key:" + i + ",value:" + map.get(i));
}
}
int dpest = 1;
List<TreeNode> leaves=new ArrayList<>();
public void dfs(TreeNode root, Map<TreeNode, TreeNode> map, int dp) {
if (root == null)
return;
if (root.left == null && root.right == null) {
//map.put(root, null);
dpest = dpest < dp ? dp : dpest;
leaves.add(root);
}
if (root.left != null) {
map.put(root.left, root);
dfs(root.left, map, dp + 1);
}
if (root.right != null) {
map.put(root.right, root);
dfs(root.right, map, dp + 1);
}
}
public void bfs(TreeNode root){
LinkedList<TreeNode> queue = new LinkedList<>();
queue.add(root);
while(!queue.isEmpty()){
TreeNode tmp=queue.poll();//queue remove
System.out.print(tmp.val+" ");
//while()
if(tmp.left!=null){
queue.add(tmp.left);//queue add
}
if(tmp.right!=null){
queue.add(tmp.right);
}
}
}
public void helper(TreeNode root, HashMap<Integer, Integer> map) {
if (root == null) {
return;
}
if (root.left == null && root.right == null) {
int val = root.val;
min = val < min ? val : min;
max = val > max ? val : max;
}
if (root.left != null) {
map.put(root.left.val, root.val);
helper(root.left, map);
}
if (root.right != null) {
map.put(root.right.val, root.val);
helper(root.right, map);
}
}
}原文地址：http://codeli.blog.51cto.com/4264850/1833651
