面試題6. 重建二叉樹

題目描述

輸入某二叉樹的前序周遊和中序周遊的結果，請重建出該二叉樹。假設輸入的前序周遊和中序周遊的結果中都不含重複的數字。例如輸入前序周遊序列{1,2,4,7,3,5,6,8}和中序周遊序列{4,7,2,1,5,3,8,6}，則重建二叉樹并傳回。

思路：

二叉樹的先序周遊中，第一個數字是根節點的值。通過根節點的值，能夠将中序周遊劃分為左子樹和右子樹兩個部分。然後遞歸左子樹和右子樹。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode reConstructBinaryTree(int[] pre,int[] in) {
        if(pre.length == 0 || in.length == 0) {
            return null;
        }
        return core(pre, 0, pre.length-1, in, 0, in.length-1);
    }
    
    public TreeNode core(int[] pre, int preStart, int preEnd, int[] in, int inStart, int inEnd) {
        
        if(preStart > preEnd || inStart > inEnd) {
            return null;
        }
        
        TreeNode root = new TreeNode(pre[preStart]);
        
        for(int i = inStart; i <= inEnd; i++) {
            if(in[i] == pre[preStart]) {
                root.left = core(pre, preStart + 1, preStart + i - inStart, in, inStart, i - 1);
                root.right = core(pre, preStart + i -inStart + 1, preEnd, in, i + 1, inEnd);
                break;
            }
        }
        return root;
    }
}

寫法2：
    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
        if (pre.length == 0) {
            return null;
        }
        return reConstruct(pre, 0, pre.length - 1, 
                           in,  0, in.length - 1);
    }
    
    public TreeNode reConstruct(int[] pre, int preStart, int preEnd, 
                                int[] in, int inStart, int inEnd) {
        if (preStart > preEnd || inStart > inEnd) {
            return null;
        }
        TreeNode node = new TreeNode(pre[preStart]);
        int headIndex = getIndex(in, node.val);
        node.left = reConstruct(pre, preStart + 1, preStart + (headIndex - inStart),
                                in, inStart, headIndex - 1);
        node.right = reConstruct(pre, preStart + 1 + (headIndex - inStart), preEnd,
                                in, headIndex + 1, inEnd);
        return node;
    }
    
    public int getIndex(int[] in, int target) {
        for (int i = 0; i < in.length; i++) {
            if (in[i] == target) {
                return i;
            }
        }
        return 0;
    }