Reconstruction of a binary tree to prove safety Offer-

Subject 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.

```public class Solution {
class TreeNode {
int val;
TreeNode left;
TreeNode right;

TreeNode(int x) {
val = x;
}
}

public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
if (pre != null && in != null)
return solution(pre, in, 0, pre.length - 1, 0, in.length - 1);
else
return null;
}

private TreeNode solution(int[] pre, int[] in, int pStart, int pEnd, int iStart, int iEnd) {

if (pStart > pEnd || iStart > iEnd)
return null;

TreeNode tree = new TreeNode(pre[pStart]);

for (int root = iStart; root <= iEnd; root++)
if (in[root] == pre[pStart]) {
int leftLen = root - iStart;
tree.left = solution(pre, in, pStart + 1, pStart + leftLen, iStart, root - 1);
tree.right = solution(pre, in, pStart + leftLen + 1, pEnd, root + 1, iEnd);
break;
}
return tree;
}
}
```

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Origin www.cnblogs.com/zty-lyq/p/11449441.html
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