leetCode 105.Construct Binary Tree from Preorder and Inorder Traversal (根据前序遍历和中序遍历构造二叉树)

标签：二叉树   leetcode   

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:

You may assume that duplicates do not exist in the tree.

思路：首先根据前序遍历得到根节点，然后在中序遍历中得到根节点的位置，左边的为左子树，右边的为右子树。

然后再递归求解左子树和右子树的构造即可。代码如下：

/**
 * 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) {
        /**
         * 1.根据前序遍历，先确定根节点
         * 2.然后在中序遍历中查找根节点，确定根节点在中序遍历的位置
         * 3.根据索引位置分割左右子树的前序和中序遍历
         * 4.递归求解根节点的左右子树
         */ 
        if(preorder.length == 0 || inorder.length == 0)
            return null;
        TreeNode root = new TreeNode(preorder[0]);
        int k = 0;
        for(; k < inorder.length; k++){
            if(inorder[k] == preorder[0]){
                break;
            }
        }
        int[] p1 = Arrays.copyOfRange(preorder,1,k+1);
        int[] q1 = Arrays.copyOfRange(preorder,k+1,preorder.length);
        int[] p2 = Arrays.copyOfRange(inorder,0,k);
        int[] q2 = Arrays.copyOfRange(inorder,k+1,inorder.length);
        
        root.left = buildTree(p1,p2);
        root.right = buildTree(q1,q2);
        return root;
    }
}

leetCode 105.Construct Binary Tree from Preorder and Inorder Traversal (根据前序遍历和中序遍历构造二叉树)

标签：二叉树   leetcode   

原文地址：http://blog.csdn.net/xygy8860/article/details/47189635