二叉搜索树的最小绝对差

给定一个所有节点为非负值的二叉搜索树，求树中任意两节点的差的绝对值的最小值。

1.采用非递归遍历二叉搜索树

 1  public void midOrderTree(Node node){
 2         if (node == null) {
 3             return;
 4         }
 5         Stack<Node> stack = new Stack<>();
 6         Node cur = node;
 7         int res = node.getVal();
 8         int temp = 0;
 9         while (cur != null || !stack.isEmpty()) {
10             while (cur != null){
11                 stack.push(cur);
12                 cur = cur.getLeft();
13             }
14             cur = stack.pop();
15             res = Math.min(cur.getVal() - temp, res);
16             temp = cur.getVal();
17             cur = cur.getRight();
18         }
19     }

2.采用递归的方式来遍历得到一个顺序的数组，然后再求最小值。

 1 private List<Integer> list = new ArrayList<>();
 2     public void midOrderTree3(Node node) {
 3         if (node == null) {
 4             return;
 5         }
 6         midOrderTree3(node.getLeft());
 7         list.add(node.getVal());
 8         midOrderTree3(node.getRight());
 9     }
10 
11     public int getMinValue(List<Integer> list){
12         if (list == null) {
13             return -1;
14         }
15         int m = 0;
16         for (int i = 1; i < list.size(); i++) {
17             m = Math.min(list.get(i) - list.get(i - 1), m);
18         }
19         return m;
20     }

标签：node,cur,getVal,int,list,二叉,最小值,搜索,null
来源： https://www.cnblogs.com/seedss/p/12995511.html