LeetCode 刷题【103. 二叉树的锯齿形层序遍历、104. 二叉树的最大深度、105. 从前序与中序遍历序列构造二叉树】

103. 二叉树的锯齿形层序遍历

自己做

解：先序遍历

java 复制代码

/**
 * 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 {
    List<List<Integer>> res = new ArrayList();
    public void preTravelTree(TreeNode root, int level){
        if(root == null)
            return;
        
        if(res.size() == level){                //添加列表【新的一层】
            List<Integer> new_level = new ArrayList();
            new_level.add(root.val);
            res.add(new_level);
        }
        else{                                   //已有列表
            List<Integer> old_level = res.get(level);
            if(level % 2 == 1){                 //奇数层，头插
                old_level.addFirst(root.val);
            }
            else{                               //偶数层，尾插
                old_level.addLast(root.val);
            }
        }

        preTravelTree(root.left, level + 1);
        preTravelTree(root.right, level + 1);
    }


    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        preTravelTree(root, 0);
        return res;
    }
}

104. 二叉树的最大深度

自己做

解：先序遍历

java 复制代码

/**
 * 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 {
    private int min_level = 0;

    public void preTravelTree(TreeNode root, int level){
        if(root == null)
            return;

        if(level > min_level)
            min_level = level;

        preTravelTree(root.left, level + 1);
        preTravelTree(root.right, level + 1);
    }

    public int maxDepth(TreeNode root) {
        preTravelTree(root, 1);
        return min_level;
    }
}

105. 从前序与中序遍历序列构造二叉树

自己做

解1：划分回溯

java 复制代码

/**
 * 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 {
    private int i = 0;

    public TreeNode inorderTravel(int[] preorder, int[] inorder, int begin, int end){
        if(begin > end)                                                         //到底了
            return null;

        TreeNode root = new TreeNode(preorder[i++]);                            //当前结点
        
        int mid = begin;                                                        //记录划分位置
        for(int j = begin; j <= end && preorder[i - 1] != inorder[mid]; j++)    //找划分位置
            mid++;

        TreeNode left = inorderTravel(preorder, inorder, begin, mid - 1);      //左子树
        TreeNode right = inorderTravel(preorder, inorder, mid + 1, end);       //右子树
        root.left = left;
        root.right =right;

        return root;
    }

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return inorderTravel(preorder, inorder, 0, preorder.length - 1);
    }
}

解2：哈希集合版

java 复制代码

/**
 * 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 {
    private int i = 0;
    Map<Integer, Integer> m = new HashMap();                                     //构建键值对集合

    public TreeNode inorderTravel(int[] preorder, int[] inorder, int begin, int end){
        if(begin > end)                                                         //到底了
            return null;

        int mid = m.get(preorder[i]);                                           //获得分割点
        TreeNode root = new TreeNode(preorder[i++]);                            //当前结点
        TreeNode left = inorderTravel(preorder, inorder, begin, mid - 1);       //左子树
        TreeNode right = inorderTravel(preorder, inorder, mid + 1, end);        //右子树
        root.left = left;
        root.right =right;

        return root;
    }

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        for(int i = 0; i < preorder.length; i++)
            m.put(preorder[i], 0);
        for(int i = 0; i < preorder.length; i++)
            m.put(inorder[i], i);
        
        return inorderTravel(preorder, inorder, 0, preorder.length - 1);
    }
}