力扣labuladong——一刷day51

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前言

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二叉树的递归分为「遍历」和「分解问题」两种思维模式，这道题需要用到「分解问题」的思维。

一、力扣106. 从中序与后序遍历序列构造二叉树

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    Map<Integer,Integer> map = new HashMap<>();
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        for(int i = 0; i < inorder.length; i ++){
            map.put(inorder[i],i);
        }
        return fun(inorder,postorder,0,inorder.length-1,0,postorder.length-1);
    }
    public TreeNode fun(int[] inorder, int[] postorder, int inLow, int inHigh,int postLow,int  postHigh){
        if(inLow > inHigh){
            return null;
        }
        int index = map.get(postorder[postHigh]);
        TreeNode cur = new TreeNode(postorder[postHigh]);
        int len = index - inLow;
        cur.left = fun(inorder,postorder,inLow,index-1,postLow,postLow+len-1);
        cur.right = fun(inorder,postorder,index+1,inHigh,postLow+len,postHigh-1);
        return cur;
    }
}

二、力扣889. 根据前序和后序遍历构造二叉树

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    Map<Integer,Integer> map = new HashMap<>();
    public TreeNode constructFromPrePost(int[] preorder, int[] postorder) {
        for(int i = 0; i < postorder.length; i ++){
            map.put(postorder[i],i);
        }
        return fun(preorder,postorder,0,preorder.length-1,0,postorder.length-1);
    }
    public TreeNode fun(int[] preorder,int[] postorder,int preLow, int preHigh,int postLow,int postHigh){
        if(preLow > preHigh){
            return null;
        }
        TreeNode cur = new TreeNode(preorder[preLow]);
        if(preLow + 1 <= preHigh){
            int index = map.get(preorder[preLow+1]);
            int len = index - postLow +1;
            cur.left = fun(preorder,postorder,preLow+1,preLow+len,postLow,index);
            cur.right = fun(preorder,postorder,preLow+len+1,preHigh,index+1,postHigh-1);
        }
        return cur;
    }
}

三、力扣331. 验证二叉树的前序序列化

class Solution {
    public boolean isValidSerialization(String preorder) {
        int edge = 1;
        for(String s : preorder.split(",")){
            if(s.equals("#")){
                edge -= 1;
                if(edge < 0){
                    return false;
                }
            }else{
                edge -= 1;
                if(edge < 0){
                    return false;
                }
                edge += 2;
            }
        }
        return edge == 0;
    }
}

四、力扣426. 将二叉搜索树转化为排序的双向链表

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val,Node _left,Node _right) {
        val = _val;
        left = _left;
        right = _right;
    }
};
*/

class Solution {
    Node pre = null, first = null;
    boolean flag = true;
    public Node treeToDoublyList(Node root) {
        if(root == null){
            return null;
        }
        fun(root);
        first.left = pre;
        pre.right = first;
        return first;
    }
    public void fun(Node root){
        if(root == null){
            return ;
        }
        fun(root.left);
        if(pre != null){
            pre.right = root;
            root.left = pre;
        }
        pre = root;
        if(flag){
            first = root;
            flag = false;
        }
        fun(root.right);
    }
}