求一颗二叉树中两个节点的最低公共父节点

题目:求一棵二叉树中两个节点的最低公共父节点

思路:递归 和 非递归

public static TreeNode getLastCommonParentRec(TreeNode root, TreeNode node1, TreeNode node2) {
        if (found(root.left, node1)) {
            if (found(root.right, node2)) {
                return root;
            } else {
                return getLastCommonParentRec(root.left, node1, node2);
            }
        } else {
            if (found(root.left, node2)) {
                return root;
            } else {
                return getLastCommonParentRec(root.right, node1, node2);
            }
        }
    }

    public static boolean found(TreeNode root, TreeNode node) {
        if (root == null || node == null)
            return false;
        if (root == node)
            return true;
        boolean found = found(root.left, node);
        if (!found) {
            found = found(root.right, node);
        }
        return found;
    }

简洁版递归:

public static TreeNode getLastCommonParentRec2(TreeNode root, TreeNode node1, TreeNode node2) {
        if (root == null)
            return null;
        if (root.equals(node1) || root.equals(node2)) {
            return root;
        }
        TreeNode left = getLastCommonParentRec2(root.left, node1, node2);
        TreeNode right = getLastCommonParentRec2(root.right, node1, node2);

        if (left != null && right != null) {
            return root;
        }

        if (left != null)
            return left;
        return right;
    }
原文地址:https://www.cnblogs.com/lfdingye/p/7372541.html