題目鏈接

tag:

• Medium；

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

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

Example:

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
Return the following binary tree:
3
/ \
9 20
?/ \
15 7

思路：
??由于先序的順序的第一個肯定是根，所以原二叉樹的根節(jié)點可以知道，題目中給了一個很關(guān)鍵的條件就是樹中沒有相同元素，有了這個條件我們就可以在中序遍歷中也定位出根節(jié)點的位置，并以根節(jié)點的位置將中序遍歷拆分為左右兩個部分，分別對其遞歸調(diào)用原函數(shù)即可。代碼如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if (preorder.size() != inorder.size()) return NULL;
        return buildTree(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);
    }
    
    TreeNode* buildTree(vector<int>& preorder, int preLeft, int preRight, vector<int>& inorder, int inLeft, int inRight) {
        if (preLeft > preRight || inLeft > inRight) return NULL;
        int i = 0;
        for (i=inLeft; i<inRight; ++i) {
            if (preorder[preLeft] == inorder[i]) break;
        }
        
        TreeNode *cur = new TreeNode(preorder[preLeft]);
        cur->left = buildTree(preorder, preLeft + 1, preLeft+i-inLeft, inorder, inLeft, i-1);
        cur->right = buildTree(preorder, preLeft+i-inLeft+1, preRight, inorder, i+1, inRight);
        return cur;
    }
};