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To prove safety offer face questions 7 (java version): reconstruction of a binary tree
Title Description
And enter the result in a preorder traversal of a binary tree in preorder traversal of the binary tree a rebuild. Suppose Results preorder traversal order and input of duplicate numbers are free. Before entering e.g. preorder traversal sequence {1,2,4,7,3,5,6,8} and {4,7,2,1,5,3,8,6} order traversal sequence, and the reconstructed binary tree return.
Thinking
- Paint is not easy to think out of bounds
- Experience recursive thought
- The title parameter passed in the recursion, the constant array is preorder traversal sequence and the first two, he said change is a value representative of the subtree starting index and ending index
the complexity
Time complexity: continuous half, the time complexity is O (logn)
Complexity Space: no use of additional memory space, the space complexity is O (1)
/**
* 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) {
// 1.健壮性判断
if(pre.length != in.length || pre.length <=0 || in.length <=0 )
return null;
// 2.开始正常执行
TreeNode root = reConstructBinaryTree(pre, 0, pre.length-1, in, 0, in.length-1);
return root;
}
public TreeNode reConstructBinaryTree(int[] pre, int preStart, int preEnd, int[] in, int inStart, int inEnd){
/*
1. 在前序遍历中找到根节点
2. 在中序遍历中找到根节点
3. 找出左子树(注意判断左子树是否存在)及其对应的前序遍历和中序遍历结果
4. 找出右子树(注意判断右子树是否存在)及其对应的前序遍历和中序遍历结果
*/
//1.
TreeNode root = new TreeNode(pre[preStart]);
//2.
int leftLen=0;
for(int i=inStart; i<= inEnd; i++){
if(in[i] == pre[preStart])
break;
leftLen++;
}
//3.
if(leftLen>0){
int preStartLeft = preStart+1;
int preEndLeft = preStart+leftLen;
int inStartLeft = inStart;
int inEndLeft = inStart + leftLen -1;
root.left = reConstructBinaryTree(pre, preStartLeft, preEndLeft, in, inStartLeft, inEndLeft);
}
if((preEnd - preStart) - leftLen > 0 ){
int preStartRight = preStart+leftLen+1;
int preEndRight = preEnd;
int inStartRight = inStart+leftLen+1;
int inEndRight = inEnd;
root.right = reConstructBinaryTree(pre, preStartRight, preEndRight, in, inStartRight, inEndRight);
}
return root;
}
}