树常见的算法操作

树是数据结构中很重要的一部分，也是各大公司面试常考部分。

继树的各种遍历算法之后，今天又整理一下树的常见算法操作。

本文包括:

1.求节点的最近公共祖先

2.树的序列化与反序列化

3.已知先序遍历和中序遍历构造二叉树

4.已知中序遍历和后序遍历构造二叉树

1.求节点最近的公共祖先

此题不同的要求有不同的解法

如果已知树中的每一个结点有指向父节点的指针：

思路:从给定节点遍历到根节点，当父节点相等时返回。

解法1

private ArrayList<TreeNode> getPath2Root(TreeNode node) {
        ArrayList<TreeNode> list = new ArrayList<TreeNode>();
        while (node != null) {
            list.add(node);
            node = node.parent;
        }
        return list;
    }
    public TreeNode lowestCommonAncestor(TreeNode node1, TreeNode node2) {
        ArrayList<TreeNode> list1 = getPath2Root(node1);
        ArrayList<TreeNode> list2 = getPath2Root(node2);

        int i, j;
        for (i = list1.size() - 1, j = list2.size() - 1; i >= 0 && j >= 0; i--, j--) {
            if (list1.get(i) != list2.get(j)) {
                return list1.get(i).parent;
            }
        }
        return list1.get(i+1);
    }

解法2:

private TreeNode getRoot(node) {
        while (node.parent != null) {
            node = node.parent;
        }
        return node;
    }

    private TreeNode getAncestor(TreeNode root, TreeNode node1, TreeNode node2) {
        if (root == null || root == node1 || root == node2) {
            return root;
        }

        TreeNode left = getAncestor(root.left, node1, node2);
        TreeNode right = getAncestor(root.right, node1, node2);

        if (left != null && right != null) {
            return root;
        }
        if (left != null) {
            return left;
        }
        if (right != null) {
            return right;
        }
        return null;
    }

    public TreeNode lowestCommonAncestor(TreeNode node1, TreeNode node2) {
        if (node1 == null || node2 == null) {
            return null;
        }
        TreeNode root = getRoot(node1);
        return getAncestor(root, node1, node2);
    }

如果树中的节点不带有指向父节点的指针:

思路利用先序遍历和中序遍历的特点，利用先序遍历分辨根节点，根据根节点和给定节点AB的位置关系来确定公共祖先，根据中序遍历结果，两个节点中间的节点即为所求祖先，若中间没有节点，则中序遍历中下标较小的即为祖先。

 TreeNode node =null;
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode A, TreeNode B) {
        // write your code here

        TreeNode result =null;
        ArrayList<TreeNode> inorder = new ArrayList<TreeNode>();
        ArrayList<TreeNode> preorder = new ArrayList<TreeNode>();
        inorderto(root,inorder);
        preorderto(root,preorder);
        //得到AB节点在中序遍历中的下标
        int aValue = inorder.indexOf(A);
        int bValue = inorder.indexOf(B);
        int lastindex = 0 ;
        for(int i=0;i<preorder.size();){
            int rootValue = inorder.indexOf(preorder.get(i));
            if((aValue<=rootValue && bValue>=rootValue)||(aValue>=rootValue && bValue<=rootValue)){
                result = preorder.get(i);
                break;
            }

            else
                i++;
        }
        return result;
    }
    public void inorderto(TreeNode root,ArrayList<TreeNode> list){
        if(root==null)
            return;
        else{
            inorderto(root.left,list);
            list.add(root);
            inorderto(root.right,list);
        }
    }
    public void preorderto(TreeNode root,ArrayList<TreeNode> list){
        if(root==null)
            return;
        else{
            list.add(root);
            preorderto(root.left,list);
            preorderto(root.right,list);
        }
    }

2.树的序列化与反序列化

将树的节点存入文件，再从文件读出构造出树称为树的序列化与反序列化。

思路:利用先序遍历(其他遍历亦可)，将树的节点值存入字符串，若节点为空则存入#，每个节点之间用‘,‘隔开用来区分数字。

读出时亦用先序遍历的方法。

 public String serialize(TreeNode root) {
        // write your code here
        String str = "";
        Stack<TreeNode> stack = new Stack<TreeNode>();
        if(root!=null)
            stack.push(root);
        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            if(node!=null){
                str += node.val + ",";
                stack.push(node.right);
                stack.push(node.left);
            }
            else
                str += "#,";
        }
        return str;
    }
    int index =0 ;
    public TreeNode deserialize(String data) {
       if(data=="")
            return null;
        String val = outValue(index,data);
        index += 1;
        if(val.equals("#"))
            return null;
        else{
            TreeNode node = new TreeNode(Integer.parseInt(val));

            node.left = deserialize(data);
            node.right = deserialize(data);
            return node;
        }
    }

    public String outValue(int index,String str){
        String res = "";
        for(int i=0,j=0;i<str.length();i++){
            if(str.charAt(i)==‘,‘)
                j++;
            else if(j==index)
                res += str.charAt(i);
            if(j>index)
                break;
        }
        return res;
    }

3.已知先序遍历和中序遍历构造二叉树

思路:利用先序遍历区分根节点，利用中序遍历区分左右子树

private int findPosition(int[] arr, int start, int end, int key) {
        int i;
        for (i = start; i <= end; i++) {
            if (arr[i] == key) {
                return i;
            }
        }
        return -1;
    }

    private TreeNode myBuildTree(int[] inorder, int instart, int inend,
            int[] preorder, int prestart, int preend) {
        if (instart > inend) {
            return null;
        }

        TreeNode root = new TreeNode(preorder[prestart]);
        int position = findPosition(inorder, instart, inend, preorder[prestart]);

        root.left = myBuildTree(inorder, instart, position - 1,
                preorder, prestart + 1, prestart + position - instart);
        root.right = myBuildTree(inorder, position + 1, inend,
                preorder, position - inend + preend + 1, preend);
        return root;
    }

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (inorder.length != preorder.length) {
            return null;
        }
        return myBuildTree(inorder, 0, inorder.length - 1, preorder, 0, preorder.length - 1);
    }

4.已知中序遍历和后序遍历构造二叉树

思路：和前者思路差不多，利用后序遍历来寻找根节点，利用中序遍历分辨左右子树

public TreeNode buildTree(int[] inorder, int[] postorder) {
        // write your code here
        if(inorder.length != postorder.length)
            return null;
        return myTree(inorder,0,inorder.length-1,postorder,0,postorder.length-1);
    }
    public TreeNode myTree(int[]inorder,int instart,int inend,
            int[] postorder,int poststart,int postend){
        if(instart>inend)
            return null;
        TreeNode root = new TreeNode(postorder[postend]);
        int position = findPosition(inorder,instart,inend,postorder[postend]);
        root.left = myTree(inorder,instart,position-1,postorder,poststart,poststart+position-1-instart);
        root.right = myTree(inorder,position+1,inend,postorder,position - inend + postend ,postend-1);
        return root;
    }
    private int findPosition(int[] arr, int start, int end, int key) {
        int i;
        for (i = start; i <= end; i++) {
            if (arr[i] == key) {
                return i;
            }
        }
        return -1;
    }