給定一棵樹的前序遍歷 preorder 與中序遍歷 inorder。請構造二叉樹並返回其根節點。

答案：

/**
 * 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 Map<Integer, Integer> map;
    public TreeNode build(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd) {
        //注意迴圈條件，先序的左邊小於右邊？--0個元素時，說明已經為空
        if(preStart > preEnd) {
            return null;
        }
        int in_root = map.get(preorder[preStart]);
        
 
int left_size = in_root - inStart;
        TreeNode root = new TreeNode(inorder[in_root]);
        root.left = build(preorder, preStart + 1, preStart + left_size, inorder, inStart, in_root - 1);
        //注意是preEnd，而不是in_root
        root.right = build(preorder, preStart + left_size + 1, preEnd, inorder, in_root + 1, inEnd);
        
 
return root;
    }
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        map = new HashMap<>();
        for(int i = 0; i < inorder.length; i++) {
            map.put(inorder[i], i);
        }
        return build(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
    }
}

本文來自部落格園，作者：劉金輝，轉載請註明原文連結：https://www.cnblogs.com/liujinhui/p/15729432.html