Construct Binary Tree from Preorder and Inorder Traversal-先序和中序构造树

最新推荐文章于 2021-07-06 08:51:48 发布

原创最新推荐文章于 2021-07-06 08:51:48 发布 · 493 阅读

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本文介绍了一种使用递归方法根据给定的先序和中序遍历构建二叉树的算法，并提供了详细的代码实现。

问题描述：Given preorder and inorder traversal of a tree, construct the binary tree.还有个是中序和后序。其实两个题本质是一样的。
问题分析：暂时只想到递归的方法来解题，可是看到耗时似乎蛮高的，不知道还有什么更搞笑的解法。
这个思路没什么好讲的，通过先序或者后序找出根节点，然后将中序的序列分开，这样就可以找出左子树和右子树，然后对左子树和右子树继续用递归即可搞定。代码如下：

<span style="font-family:Microsoft YaHei;">/**
 * 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 ||inorder == null || preorder.length == 0 || preorder.length != inorder.length)
            return null;
        return buildTree(preorder,inorder, 0,preorder.length-1, 0, inorder.length-1);
    }
    public TreeNode buildTree(int[] preorder, int[] inorder, int i, int j, int k, int l){
        if(i == j)
            return new TreeNode(preorder[i]);
        int index = k;
        for(; index <=l && preorder[i] != inorder[index]; index++);
        // for(int idx = k+1; idx <= l; idx++)
        //     if(preorder[i] == inorder[idx]){
        //         index = idx;
        //         break;
        //     }
        TreeNode root = new TreeNode(preorder[i]);
        if(index > k)
            root.left = buildTree(preorder, inorder, i+1, i + index-k, k, index -1);
        if(index < l)
            root.right = buildTree(preorder, inorder, i+index-k + 1, j, index +1, l);
        return root;
    }
}</span>

<span style="font-family:Microsoft YaHei;">/**
 * 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 ||inorder == null || preorder.length == 0 || preorder.length != inorder.length)
            return null;
        return buildTree(preorder,inorder, 0,preorder.length-1, 0, inorder.length-1);
    }
    public TreeNode buildTree(int[] preorder, int[] inorder, int i, int j, int k, int l){
        if(i == j)
            return new TreeNode(preorder[i]);
        int index = k;
        for(; index <=l && preorder[i] != inorder[index]; index++);
        // for(int idx = k+1; idx <= l; idx++)
        //     if(preorder[i] == inorder[idx]){
        //         index = idx;
        //         break;
        //     }
        TreeNode root = new TreeNode(preorder[i]);
        if(index > k)
            root.left = buildTree(preorder, inorder, i+1, i + index-k, k, index -1);
        if(index < l)
            root.right = buildTree(preorder, inorder, i+index-k + 1, j, index +1, l);
        return root;
    }
}</span>