前序遍历和中序遍历树构造二叉树

 

73. 前序遍历和中序遍历树构造二叉树

根据前序遍历和中序遍历树构造二叉树.

样例

给出中序遍历：[1,2,3]和前序遍历：[2,1,3]. 返回如下的树:

  2
 / \
1   3

注意事项

你可以假设树中不存在相同数值的节点

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param inorder: A list of integers that inorder traversal of a tree
     * @param postorder: A list of integers that postorder traversal of a tree
     * @return: Root of a tree
     */
    TreeNode * buildTree(vector &preorder, vector &inorder) {
        // write your code here
        if(preorder.size()<1 || inorder.size()<1 || preorder.size()!=inorder.size()) return NULL;
   
        TreeNode* root=new  TreeNode(preorder[0]);
        int mid=preorder[0];
        int pos=0;
        for(int i=0;i preo,prei;
        vector ino,ini;
        for(int i=1;i<=pos;i++)
        {
            preo.push_back(preorder[i]);
        }
        for(int i=pos+1;ileft=buildTree(preo,ino);
        root->right=buildTree(prei,ini);
        return root;
        
    }
};

///////////////////////////////////////////////////////////
"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
    """
    @param inorder: A list of integers that inorder traversal of a tree
    @param postorder: A list of integers that postorder traversal of a tree
    @return: Root of a tree
    """
    def buildTree(self, preorder, inorder):
        # write your code here
        if len(preorder)<1 or len(inorder)<1 or len(preorder)!=len(inorder):
            return None
        mid=preorder[0]
        pos=0
        for i in range(len(inorder)):
            if mid==inorder[i]:
                pos=i;
                break;
        prep=preorder[1:pos+1:1]
        prei=preorder[pos+1::]
        inp=inorder[:pos:]
        ini=inorder[pos+1::]
        root=TreeNode(mid)
        root.left=self.buildTree(prep,inp)
        root.right=self.buildTree(prei,ini)
        return root