106 Construct Binary Tree from Inorder and Postorder Traversal

public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (inorder == null || postorder == null || inorder.length != postorder.length || inorder.length == 0) {
            return null;
        }
        int length = inorder.length;
        int rootVal = postorder[length - 1];
        int mid = 0;
        for (int i = 0; i < length; i++) {
            if (inorder[i] == rootVal) {
                mid = i;
                break;
            }
        }
        
        int[] leftIn = Arrays.copyOfRange(inorder, 0, mid);
        int[] rightIn = Arrays.copyOfRange(inorder, mid + 1, length);
        int[] leftPost = Arrays.copyOfRange(postorder, 0, mid);
        int[] rightPost = Arrays.copyOfRange(postorder, mid, length - 1);
        TreeNode root = new TreeNode(rootVal);
        root.left = buildTree(leftIn, leftPost);
        root.right = buildTree(rightIn, rightPost);
        return root;
    }
}

知道copyOfRange(int[] array, int start, int end)方法，start和end要写准。

转载于:https://www.cnblogs.com/77rousongpai/p/4513605.html