java 105.从预订和顺序遍历构建二叉树（＃）。java

Posted 2021-05-08

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder == null || preorder.length == 0 
            || inorder == null || inorder.length == 0) return null;
        
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int i = 0; i < inorder.length; i++) {
            map.put(inorder[i], i);
        }
        
        return helper(preorder, 0, preorder.length - 1, map, 0);
    }
    
    private TreeNode helper(int[] preorder, int pl, int pr, Map<Integer, Integer> map, int il) {
        if (pl > pr) return null;
        
        TreeNode node = new TreeNode(preorder[pl]);
        if (pl == pr) {
            return node;
        }

        int i = map.get(preorder[pl]);
        int len = i - il;
        node.left = helper(preorder, pl + 1, pl + len, map, il);
        node.right = helper(preorder, pl + len + 1, pr, map, i + 1);
        return node;
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int pre = 0;
    int in = 0;
    
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return builder(preorder, inorder, Integer.MAX_VALUE);
    }
    
    private TreeNode builder(int[] preorder, int[] inorder, int border) {
        if (in >= inorder.length || inorder[in] == border) { return null; }
        
        TreeNode root = new TreeNode(preorder[pre]);
        pre++;
        root.left = builder(preorder, inorder, root.val);
        in++;
        root.right = builder(preorder, inorder, border);
        
        return root;
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private TreeNode helper(int[] preorder, int[] inorder, int rootIdx, int start, int end) {
        if (rootIdx > preorder.length - 1 || start > end) return null;
        TreeNode root = new TreeNode(preorder[rootIdx]);
        int inIdx = 0;
        for (int i = start; i <= end; i++) {
            if(inorder[i] == root.val) {
                inIdx = i;
            }
        }
        root.left = helper(preorder, inorder, rootIdx + 1, start, inIdx - 1);
        root.right = helper(preorder, inorder, rootIdx + inIdx - start + 1, inIdx + 1, end);
        return root;
    }
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return helper(preorder, inorder, 0, 0, preorder.length - 1);
    }
}