【LeetCode】Construct Binary Tree from Preorder and Inorder Traversal 解題報告，leetcodepreorder

【LeetCode】Construct Binary Tree from Preorder and Inorder Traversal 解題報告，leetcodepreorder

【原題】

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

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

【解析】

題意：根據二叉樹先序遍曆和中序遍曆的結果，構造二叉樹。跟 根據中序遍曆和後序遍曆結果構造二叉樹 類似。

先序遍曆：root - left - right，中序遍曆：left - root - right。

不同的是，先序遍曆的第一個元素就是根節點。

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    int[] preorder;    int[] inorder;        public TreeNode buildTree(int[] preorder, int[] inorder) {        if (preorder.length < 1 || inorder.length < 1) return null;        this.preorder = preorder;        this.inorder = inorder;        return getRoot(0, inorder.length - 1, 0);    }        // sub-tree range form begin to end in inorder[]    // rootpos refer the root postion in preorder[]    public TreeNode getRoot(int begin, int end, int rootpos) {        if (begin > end) return null;        TreeNode root = new TreeNode(preorder[rootpos]);        int i;        for (i = begin; i <= end; i++) {            if (inorder[i] == preorder[rootpos]) break;        }        // left part length: i - begin, right part length: end - i        root.left = getRoot(begin, i - 1, rootpos + 1);        root.right = getRoot(i + 1, end, rootpos + 1 + (i - begin));        return root;    }}